Chapter 5 Chemical Reactions and Equations

Chapter 5 Chemical Reactions and Equations 5.1 (a) single-displacement reaction; (b) anhydrous; (c) molecular equation; (d) decomposition reaction; (e) balanced equation; (f) reactant; (g) spectator ion; (h) combustion reaction; (i) precipitate 5.2 (a) combination reaction; (b) activity series; (c) chemical equation; (d) product; (e) ionic equation; (f) neutralization reaction; (g) precipitation reaction; (h) net ionic equation; (i) double-displacement reaction 5.3 Substances that are formed as a result of the reaction are the products. This reaction produces solid aluminum oxide, Al 2 O 3 (s). You should pay attention to the substance formed and the state of the material if it is given. The reactants are the substances used to form the products. The reactants are aluminum metal, Al(s), and oxygen gas, O 2 (g) (i.e. Aluminum metal reacts with oxygen gas ). 5.4 Substances that are formed as a result of the reaction are the products. Nitrogen gas and solid iodine, N 2 (g) and I 2 (s), and energy are the reaction products (i.e. Nitrogen gas and solid iodine form in an explosive reaction ). You should pay attention to the substance formed and the state of the material if it is given. The reactants are the substances used to form the products. The reactant is nitrogen triiodide, NI 3 (i.e. when nitrogen triiodide is detonated ). 5.5 Image A represents the reactants because it contains both fluorine and xenon. Images B-D show product molecules, but only image C represents the products. In a chemical reaction, mass must be conserved. In image A there are two xenon atoms and eight fluorine atoms. In order for mass to be conserved, the same number of each type of atom must be present in the products. 5.6 Make sure you look carefully at the images to distinguish between the chemical formulas for ozone (three red spheres) and oxygen (two red spheres). Image C represents the reactants because it contains both ozone and carbon monoxide. The products are represented by image D. Images A and B represent mixtures of reactants and products. 5.7 Mass must be conserved in a chemical reaction. In order for mass to be conserved, the same number of each type of atom must be present in both the products and reactants. The image is not accurate because mass is not conserved. To fix the image, you first need to count the number of each type of atom. Here s a summary of the atoms present in the reactants and products: Atom H 10 12 N 4 4 The numbers of hydrogen atoms do not match in the reactants and products. If you add one hydrogen molecule to the reactant image, the numbers will balance and the molecular-level diagram will be accurate. 5.8 Mass must be conserved in a chemical reaction. In order for mass to be conserved, the same number of each type of atom must be present in both the products and reactants. The image is not accurate because mass is not conserved. To fix the image, you first need to count the number of each type of atom. Here s a summary of the atoms present in the reactants and products: Atom H 10 10 O 6 5 5 1

The numbers of oxygen atoms do not match in the reactants and products. If you add one hydrogen molecule to the reactant image and one molecule of water (H 2 O) to the product image, the numbers will balance and the molecular-level diagram will be accurate. Atom H 12 12 O 6 6 5.9 The law of conservation of mass is obeyed when the same number of each type of atom is present in both the products and reactants. In the product image you could have 5 molecules and one unreacted atom. 5.10 The law of conservation of mass is obeyed when the same number of each type of atom is present in both the products and reactants. In the product image you could have 1 unreacted molecule and 5 unreacted atoms. Two molecules of product should be shown. 5.11 One oxygen molecule (two oxygen atoms) and two sulfur dioxide molecules must react for every two sulfur trioxide molecules (SO 3 ) produced. Since there are four sulfur dioxide molecules, both oxygen molecules will react. This means that none of the reactants remain after the reaction, and we should show four sulfur trioxide molecules as the reaction products. 5 2

5.12 One oxygen molecule (two oxygen atoms) and two NO molecules must react for every two NO 2 produced. Since there are six NO molecules, all of the oxygen molecules will react. We should show six NO 2 molecules as the reaction products. 5.13 Three signs that a chemical reaction is occurring are evident: (1) a brown gas is forming, (2) bubbles are being produced, and (3) the solution is changing color (assuming that the liquid was not green to start with). 5.14 Two signs that a chemical reaction is occurring are evident: (1) the color of the mixture is different than the color of either solution and (2) a new substance (the yellow solid) is forming. You can tell a solid is formed because the solution becomes opaque (i.e. you can t see through it). 5.15 We are starting and ending with CO 2, so no new substances are formed. A chemical reaction has not taken place. What we are observing is a change in the physical state of CO 2. 5.16 There are several signs that a chemical reaction has taken place: the formation of a blue solution indicates that a color change occurred. In addition, formation of a black deposit indicates the production of a new substance. 5.17 There is a rearrangement of the atoms into new substances. Since new substances are formed, a chemical reaction has taken place. 5.18 In the image on the left, the molecules are composed of one red and two white spheres, just as they are in the image on the right. When a chemical reaction takes place new substances are formed. A chemical reaction is not occurring here, because no new substance is formed. 5.19 The solution shown on the left has the spheres placed closer together, while in the solution shown on the right the spheres have moved farther apart. Since no new chemical substances have formed, a chemical reaction has not taken place. This is what you would expect to see in dilution. 5.20 A rearrangement of the atoms in a substance is an indication that a chemical reaction has taken place. Since the molecules have different shapes in the reactant and product images, we conclude that a chemical reaction has occurred. 5.21 A chemical equation is a chemist s shorthand way of showing what happens during a chemical reaction. It identifies the formulas of the reactants and products and demonstrates how mass is conserved during the reaction. You can think of a chemical equation as a chemist s recipe for making new substances. The reactants are the ingredients (i.e. melted chocolate or solid chocolate chips) and the products are what are produced by the reaction (i.e. fudge or cookies). Like a recipe, the chemical equation indicates the states of the materials and the quantities of substances that are used (the reactants) and produced (the products). 5.22 A chemical reaction is any process that produces a new substance. A chemical equation is the shorthand that chemists use to describe that reaction. The chemical equation includes the quantities and physical states of the substances used (the reactants) and produced (the products) in the ideal reaction. 5 3

5.23 To decide if a reaction has taken place, we have to determine whether a new substance has been formed. (a) This equation represents a chemical reaction. Carbonic acid, H 2 CO 3 (aq), has been formed from the reaction of carbon dioxide (CO 2 ) and water (H 2 O). (b) This equation represents a physical change. The chemical compositions of solid water (ice) and liquid water are the same, so a chemical reaction did not occur. (c) This equation represents a chemical reaction. Even though the same atoms are present in the reactant and product, the order in which they are connected is different so the substances are different compounds. 5.24 (a) This equation represents a chemical reaction. New substances, TiO 2 (s) and HCl(aq), are formed from TiCl 4 and H 2 O. (b) This equation represents a chemical reaction. Solid Li 2 O forms from the elements Li and O 2. (c) The process of dissolving a salt in water is usually considered a physical change. Salts are composed of ions (for example, KBr(s) is actually a collection of K + and Br ions). When a salt dissolves, the ions are separated from each other by water molecules. However, the structure of the ions does not change. 5.25 Balancing a chemical equation demonstrates how mass is conserved during a reaction. A balanced equation is a quantitative tool for determining the amounts of reactant(s) used and of product(s) produced. 5.26 Chemical equations are descriptions of particular chemical reactions. Changing subscripts in the formulas of reactant or product species changes the identities of the substances involved in the reaction the equation is describing. 5.27 In balancing the equation, we add coefficients so that the number of each type of atom is the same on the reactant and product sides of the equation. (a) To start with, we write the skeletal equation for the reaction: NaH(s) + H 2 O(l) H 2 (g) + NaOH(aq) When we count the number of each type of atom in the reactants and products, we find that this equation is already balanced. 1 Na 3 H 1 O 1 Na 3 H 1 O (b) We start by writing the skeletal equation: Al(s) + Cl 2 (g) AlCl 3 (s) Unbalanced We count the number of each type of atom on each side of the unbalanced equation to determine where to start balancing the equation. 1 Al 2 Cl 1 Al 3 Cl The chlorines are not balanced. To balance, we need to find a common factor. The smallest factor would be 6. We use coefficients of 3 and 2 on Cl 2 (g) and AlCl 3 (s), respectively Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1 Al 6 Cl 2 Al 6 Cl Now we balance the Al on the reactant side, using a coefficient of 2. 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 5 4

2 Al 6 Cl 2 Al 6 Cl Since the number of each type of atom on the reactant and product sides of the equation is the same, the equation is balanced. 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 5.28 In balancing the equation, we add coefficients so that the number of each type of atom is the same on the reactant and product sides of the equation. (a) We start by writing the skeletal equation: Na(s) + O 2 (g) Na 2 O(s) Unbalanced We count the number of each type of atom on each side of the unbalanced equation to determine where to start balancing the equation. 1 Na 2 O 2 Na 1 O You can start by balancing either the oxygen or the sodium. If you start with oxygen, you need a coefficient of 2 in front of the Na 2 O. Na(s) + O 2 (g) 2Na 2 O(s) Unbalanced 1 Na 2 O 4 Na 2 O Since there are four sodium atoms on the right, we use a coefficient of 4 with Na: 4Na(s) + O 2 (g) 2Na 2 O(s) 4 Na 2 O 4 Na 2 O Since the number of each type of atom on the reactant and product sides of the equation is the same, the equation is balanced. 4Na(s) + O 2 (g) 2Na 2 O(s) (b) We start by writing the skeletal equation: Cu(NO 3 ) 2 (s) CuO(s) + NO 2 (g) + O 2 (g) Unbalanced We count the number of each type of atom on each side of the unbalanced equation to determine where to start balancing the equation. 1 Cu 2 N 6 O 1 Cu 1 N 5 O Since copper is already balanced, we need to start with either the nitrogen or the oxygen. It is usually best to start by balancing the atoms that appear in the smallest number of compounds. In this case that is nitrogen. We balance the nitrogen by placing a coefficient of 2 in front of NO 2. Cu(NO 3 ) 2 (s) CuO(s) + 2NO 2 (g) + O 2 (g) When you count atoms, don t forget that you have also increased the number of oxygen atoms in the product at the same time you increased the number of nitrogen atoms. 1 Cu 2 N 6 O 1 Cu 2 N 7 O 5 5

Since there are not enough oxygen atoms on the reactant side (and there is only one reactant) we need to add a coefficient of 2 in front of the Cu(NO 3 ) 2 (s). We will also have to increase the coefficient for NO 2 to keep the number of nitrogen atoms balanced. 2 Cu(NO 3 ) 2 (s) CuO(s) + 4 NO 2 (g) + O 2 (g) 2 Cu 4 N 12 O 1 Cu 4 N 11 O The last step is to balance number of copper and oxygen atoms: 2 Cu(NO 3 ) 2 (s) 2 CuO(s) + 4 NO 2 (g) + O 2 (g) 2 Cu 4 N 12 O 2 Cu 4 N 12 O Since the number of each type of atom on the reactant and product sides of the equation is the same, the equation is balanced. 2 Cu(NO 3 ) 2 (s) 2 CuO(s) + 4 NO 2 (g) + O 2 (g) 5.29 There are two nitrogen molecules (N 2 ) and six chlorine molecules (Cl 2 ) in the reactant image, and 4 NCl 3 molecules in the product image. The large spacing between the molecules allows us to assume that the reactants and products are in the gas state. The balanced chemical equation is: 2N 2 (g) + 6Cl 2 (g) 4NCl 3 (g) The coefficients of a balanced equation should be written as the smallest possible whole numbers. Since each of the coefficients is divisible by 2, we reduce them by dividing each by 2. N 2 (g) + 3Cl 2 (g) 2NCl 3 (g) 5.30 There are three methane molecules (CH 4 ) and six oxygen molecules (O 2 ) in the reactant image and three carbon dioxide molecules (CO 2 ) and six water (H 2 O) molecules in the product image. The large spacing between the molecules allows us to assume the reactants and products are in the gas state. The balanced chemical equation is: 3CH 4 (g) + 6O 2 (g) 3CO 2 (g) + 6H 2 O(g) The coefficients of a balanced equation should be written as the smallest possible whole numbers. Since each of the coefficients is divisible by 3, we reduce them by dividing each by 3. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) 5.31 Magnesium is a solid and it burns in the presence of oxygen, O 2 (g), from the air. The product is also a solid (indicated by the statement ash-like substance ). Image B matches the reactants most closely because it shows the correct states of materials in the reactants. Also, if we assume that magnesium oxide is formed, the chemical formula for the product would be MgO(s). This also matches image B. The skeletal equation for this reaction is: Mg(s) + O 2 (g) MgO(s) Unbalanced 1 Mg 2 O 1 Mg 1 O We add a coefficient of 2 to MgO to balance the oxygen: Mg(s) + O 2 (g) 2MgO(s) 1 Mg 2 O 2 Mg 2 O 5 6

Then we add a coefficient of 2 to Mg to complete the balancing process: 2Mg(s) + O 2 (g) 2MgO(s) 2 Mg 2 O 2 Mg 2 O 2Mg(s) + O 2 (g) 2MgO(s) 5.32 Image A matches this description most closely. The water molecules (shown in the background of the reactant and product images) are closely spaced and randomly oriented. This is what we would expect for a substance in the liquid phase. Metallic sodium should be a solid with a well defined structure. The products include aqueous sodium hydroxide, so we expect to find Na + and OH ions surrounded by water molecules. Finally, we see several H 2 molecules in the solution. The skeletal equation for the reaction is: Na(s) + H 2 O(l) NaOH(aq) + H 2 (g) Unbalanced 1 Na 2 H 1 O 1 Na 3 H 1 O It is easiest to balance this reaction using a fractional coefficient (1/2) for the hydrogen molecule. Na(s) + H 2 O(l) NaOH(aq) + ½H 2 (g) 1 Na 2 H 1 O 1 Na 2 H 1 O This technique is best applied when everything except atoms of one element (i.e. H 2, O 2, S 8, etc.) is balanced. By using a fractional coefficient, we can change the amount of one type of atom without affecting the others. We remove the fractional coefficient in this case by multiplying all the coefficients by 2 (the denominator of the fraction) to remove the fraction. 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) 5.33 According to the balanced chemical equation for this reaction, for each molecule of H 2 and I 2 that are consumed in the reaction, two HI molecules are produced. Since three H 2 and three I 2 molecules are shown, we must show six molecules of HI in the product image of the molecular-level diagram. 5.34 According to the balanced chemical equation for this reaction, for every two molecules of NO and CO that are consumed in the reaction, one molecule of N 2 and two molecules of CO 2 are produced. Since four NO and four CO molecules are shown in the left image, we must show two molecules of N 2 and four molecules of CO 2 in the product image. 5 7

5.35 Our molecular-level diagram should include two molecules of NO 2 as the reactants and one molecule of N 2 O 4 as the products. The law of conservation of mass must be satisfied for the diagram to be accurate. In each image there are two nitrogen atoms and four oxygen atoms. 5.36 Our molecular-level diagram should include two molecules of F 2 and two molecules of H 2 O as the reactants and four HF molecules and two O 2 molecules as the products. The law of conservation of mass must be obeyed for the diagram to be accurate. In each image there are four fluorine atoms, four hydrogen atoms, and two oxygen atoms. 5.37 (a) We start by writing the skeletal equation: Al(s) + Cl 2 (g) AlCl 3 (s) Unbalanced Start by taking an inventory of atoms present on each side to help determine where to start balancing the equation. 1 Al 2 Cl 1 Al 3 Cl Only the chlorines are not balanced. To balance, we need to find a common factor for 2 and 3. The smallest factor would be 6. We use coefficients of 3 and 2 on Cl 2 (g) and AlCl 3 (s), respectively Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1 Al 6 Cl 2 Al 6 Cl We recheck our inventory and find that we need to balance the Al on the reactant side using a coefficient of 2. 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 2 Al 6 Cl 2 Al 6 Cl Since the number of each type of atom on the reactant and product sides of the equation is the same, the equation is balanced. 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) (b) We start by writing the skeletal equation: 5 8

Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + KNO 3 (aq) Unbalanced In the inventory it is perferable to keep polyatomic ions together as groups. This works when the polyatomic ions are not changed in the reaction: 1 Pb 2 NO 3 2 K 1 CrO 4 1 Pb 1 NO 3 1 K 1 CrO 4 Start by balancing nitrates or potassium ions. Using a two in front of KNO 3 balances both potassiums and nitrates! Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2KNO 3 (aq) 1 Pb 2 NO 3 2 K 1 CrO 4 1 Pb 1 NO 3 2 K 1 CrO 4 (c) We start by writing the skeletal equation: Li(s) + H 2 O(l) LiOH(aq) + H 2 (g) Unbalanced 1 Li 2 H 1 O 1 Li 3 H 1 O Notice that the hydrogens are the only atoms not balanced, but on the product side, the hydrogens appear in two different substances. If we change the number of H 2 and H 2 O molecules, the reaction will continue to be unbalanced. The solution is to use a coefficient of 2 on the H 2 O and see where it might lead you. Li(s) + 2H 2 O(l) LiOH(aq) + H 2 (g) Unbalanced 1 Li 4 H 2 O 1 Li 3 H 1 O Next balance the oxygens: Li(s) + 2H 2 O(l) 2LiOH(aq) + H 2 (g) Unbalanced 1 Li 4 H 2 O 2 Li 4 H 2 O The last step is to balance the lithium atoms and check. 2Li(s) + 2H 2 O(l) 2LiOH(aq) + H 2 (g) 2 Li 4 H 2 O 2 Li 4 H 2 O (d) We start by writing the skeletal equation: C 6 H 14 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Unbalanced 6C 14H 2O 1C 2H 3O 5 9

In the combustion reactions of hydrocarbons it is usually easiest to begin by balancing the carbon and leave the balancing of oxygen for the last step. This follows the general concepts of balancing the atoms that appear in the smallest number of substances first (you could have also started with hydrogen) and balancing the substances appearing as pure elements last. C 6 H 14 (g) + O 2 (g) 6CO 2 (g) + H 2 O(g) Unbalanced 6C 14H 2O 6C 2H 13O C 6 H 14 (g) + O 2 (g) 6CO 2 (g) + 7H 2 O(g) Unbalanced 6C 14H 2O 6C 14H 19O To balance the oxygen, we use a coefficient of 19/2 for O 2 : C 6 H 14 (g) + 19/2 O 2 (g) 6CO 2 (g) + 7H 2 O(g) While this equation is balanced, we need to remove the fraction by multiplying the entire equation by a factor of two. 2C 6 H 14 (g) + 19 O 2 (g) 12CO 2 (g) + 14H 2 O(g) 12C 28H 38O 12C 28H 38O 5.38 (a) You are given the skeletal equation: Mg(s) + O 2 (g) MgO(s) Unbalanced Start by taking an inventory of atoms present on each side to help determine where to start balancing the equation. 1 Mg 2 O 1 Mg 1 O Since the oxygens are not balanced, we use a coefficient of 2 in front of MgO Mg(s) + O 2 (g) 2MgO(s) Unbalanced 1 Mg 2 O 2 Mg 2 O Coefficient of 2 on Mg balances the equation: 2Mg(s) + O 2 (g) 2MgO(s) 2 Mg 2 O 2 Mg 2 O 5 10

(b) Zn(s) + 2AgNO 3 (aq) Zn(NO 3 ) 2 (aq) + 2Ag(s) Zn(s) + AgNO 3 (aq) Zn(NO 3 ) 2 (aq) + Ag(s) Unbalanced In the inventory it is perferable to keep polyatomic ions together as groups. This works when the polyatomic ions are not changed in the reaction: 1 Zn 1 Ag 1 NO 3 1 Zn 1 Ag 2 NO 3 Since the nitrates are not balanced, we start by using a coefficient of 2 in front of AgNO 3 Zn(s) + 2AgNO 3 (aq) Zn(NO 3 ) 2 (aq) + Ag(s) 1 Zn 2 Ag 2 NO 3 1 Zn 1 Ag 2 NO 3 A coefficient in front of the Ag in the product finish the balancing process. Zn(s) + 2AgNO 3 (aq) Zn(NO 3 ) 2 (aq) + 2Ag(s) 1 Zn 2 Ag 2 NO 3 1 Zn 2 Ag 2 NO 3 (c) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) C 2 H 2 (g) + O 2 (g) CO 2 (g) + H 2 O(g) unbalanced 2 C 2 H 2 O 1 C 2 H 3 O In the combustion reactions of hydrocarbons it is usually easiest to begin by balancing the carbon and leave the balancing of oxygen for the last step. This follows the general concepts of balancing the atoms that appear in the smallest number of substances first (you could have also started with hydrogen) and balancing the substances appearing as pure elements last. Start with a coefficient of 2 in front of CO 2. C 2 H 2 (g) + O 2 (g) 2CO 2 (g) + H 2 O(g) unbalanced 2 C 2 H 2 O 2 C 2 H 5 O To balance the oxygen, we use a coefficient of 5/2 for C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O(g) unbalanced 2 C 2 H 5 O 2 C 2 H 5 O While this equation is balanced, we need to remove the fraction by multiplying the entire equation by a factor of two. 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) 4 C 4 H 10 O 4 C 4 H 10 O (d) C 12 H 22 O 11 (s) + 8KClO 3 (l) 12CO 2 (g) + 11H 2 O(g) + 8KCl(s) C 12 H 22 O 11 (s) + KClO 3 (l) CO 2 (g) + H 2 O(g) + KCl(s) Unbalanced 12C 22H 14O 1K 1Cl 1C 2H 3O 1K 1Cl Start by balaning products with the reactant C 12 H 22 O 11. As with hydrocarbons, balance C and H first: C 12 H 22 O 11 (s) + KClO 3 (l) 12CO 2 (g) + 11H 2 O(g) + KCl(s) 12C 22H 14O 1K 1Cl 12C 22H 35O 1K 1Cl There are 35 oxygen atoms in the products and 11 in the reactants. Ignore the oxygens in KClO 3 because it is unbalanced. This means that KClO 3 needs to provide 24 oxgyen atoms. A coefficient of eight will provide those oxygens. 5 11

C 12 H 22 O 11 (s) + 8KClO 3 (l) 12CO 2 (g) + 11H 2 O(g) + KCl(s) 12C 22H 35O 8K 8Cl 12C 22H 35O 1K 1Cl A coefficient of eight in front of KCl balances the equation: C 12 H 22 O 11 (s) + 8KClO 3 (l) 12CO 2 (g) + 11H 2 O(g) + 8KCl(s) 12C 22H 35O 8K 8Cl 12C 22H 35O 8K 8Cl 5.39 (a) We start by writing the skeletal equation: CuCl 2 (aq) + AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + AgCl(s) Unbalanced When balancing chemical equations that contain polyatomic ions which do not undergo chemical change during the reaction, treat the ions as whole units in the same way that you would treat individual atoms. 1 Cu 2 Cl 1 Ag 1 NO 3 1 Cu 1 Cl 1 Ag 1 NO 3 You can balance this equation by starting with either Cl or NO 3. CuCl 2 (aq) + AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2AgCl(s) 1 Cu 2 Cl 1 Ag 1 NO 3 1 Cu 2 Cl 2 Ag 2 NO 3 CuCl 2 (aq) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2AgCl(s) 1 Cu 2 Cl 2 Ag 2 NO 3 1 Cu 2 Cl 2 Ag 2 NO 3 Since the number of each type of atom on both sides of the equation is the same, the equation is balanced. (b) We start by writing the skeletal equation: S 8 (s) + O 2 (g) SO 2 (g) Unbalanced 8 S 2 O 1 S 2 O Begin by balancing S, because O already appears to be balanced. S 8 (s) + O 2 (g) 8SO 2 (g) 8 S 2 O 8 S 16 O Add a coefficient of 8 for O 2 to balance the equation. S 8 (s) + 8O 2 (g) 8SO 2 (g) 8 S 16 O 8 S 16 O Since the number of each type of atom on both sides of the equation is the same, the equation is balanced. 5 12

(c) We start by writing the skeletal equation: C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Unbalanced 3 C 8 H 2 O 1 C 2 H 3 O In combustion reactions of hydrocarbons (hydrocarbon + molecular oxygen carbon dioxide + water), we begin by balancing the carbon. This follows the general concept of balancing the atoms that appear in the smallest number of substances first (you could have also started with hydrogen). We add a coefficient of 3 for CO 2. C 3 H 8 (g) + O 2 (g) 3CO 2 (g) + H 2 O(g) 3 C 8 H 2 O 3 C 2 H 7 O Continue balancing all the other types of atoms in C 3 H 8 before proceeding to balance the oxygen. We add a coefficient of 4 for H 2 O. C 3 H 8 (g) + O 2 (g) 3CO 2 (g) + 4H 2 O(g) 3 C 8 H 2 O 3 C 8 H 10 O Finally, add a coefficient of 5 for O 2. C 3 H 8 (l) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) 3 C 8 H 10 O 3 C 8 H 10 O Since the number of each type of atom on both sides of the equation is the same, the equation is balanced. 5.40 (a) Al 2 O 3 (s) + 6HCl(aq) 2AlCl 3 (aq) + 3H 2 O(l) Al 2 O 3 (s) + HCl(aq) AlCl 3 (aq) + H 2 O(l) Unbalanced 2 Al 3 O 1 H 1 Cl 1 Al 1 O 2 H 1 Cl We begin by balancing the Al with coefficient of 2 for AlCl 3. Al 2 O 3 (s) + HCl(aq) 2AlCl 3 (aq) + H 2 O(l) 2 Al 3 O 1 H 1 Cl 2 Al 1 O 2 H 6 Cl Continue balancing all the types of atoms in Al 2 O 3 before proceeding to balance the other elements. Al 2 O 3 (s) + HCl(aq) 2AlCl 3 (aq) + 3H 2 O(l) 2 Al 3 O 1 H 1 Cl 2 Al 3 O 6 H 6 Cl Balance the H and Cl by adding a coefficient of 6 for HCl. 5 13

Al 2 O 3 (s) + 6HCl(aq) 2AlCl 3 (aq) + 3H 2 O(l) 2 Al 3 O 6 H 6 Cl 2 Al 3 O 6 H 6 Cl Since the number of each type of atom on both sides of the equation is the same, the equation is balanced. (b) PCl 3 (l) + 3AgF(s) PF 3 (g) + 3AgCl(s) PCl 3 (l) + AgF(s) PF 3 (g) + AgCl(s) Unbalanced 1 P 3 Cl 1 Ag 1 F 1 P 1 Cl 1 Ag 3 F Begin by balancing either Cl or F. PCl 3 (l) + AgF(s) PF 3 (g) + 3AgCl(s) 1 P 3 Cl 1 Ag 1 F 1 P 3 Cl 3 Ag 3 F PCl 3 (l) + 3AgF(s) PF 3 (g) + 3AgCl(s) 1 P 3 Cl 3 Ag 3 F 1 P 3 Cl 3 Ag 3 F Since the number of each type of atom on both sides of the equation is the same, the equation is balanced. (c) 2NO 2 (g) 2NO(g) + O 2 (g) NO 2 (g) NO(g) + O 2 (g) Unbalanced 1 N 2 O 1 N 3 O Since nitrogen is balanced, we have to start with O. Since we only need half of the oxygen provided by O 2, we use a coefficient of 1/2 for O 2 to balance the equation. NO 2 (g) NO(g) + 1/2O 2 (g) 1 N 2 O 1 N 2 O Since the number of each type of atom on both sides of the equation is the same, theoretically, the reaction is balanced. However, the coefficients of a balanced equation need to be whole numbers so we multiply all the coefficients by 2 to remove the fraction. 2NO 2 (g) 2NO(g) + O 2 (g) 2 N 4 O 2 N 4 O 5.41 Copper and silver, Cu and Ag, are metals, so they, like virtually all the metals, occur in the solid state. Silver nitrate is an ionic compound. The formulas for silver ion and nitrate ion are Ag + and NO 3, respectively (see Figures 3.12 and 3.17 if you need help with the ion formulas). The formula for an aqueous solution of silver nitrate is AgNO 3 (aq). From the ion name, copper(ii), we know that the formula 5 14

for copper ion is Cu 2+. The formula for an aqueous solution of copper(ii) nitrate is Cu(NO 3 ) 2 (aq). The unbalanced chemical equation is: Cu (s) + AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + Ag(s) Unbalanced When balancing chemical equations that contain polyatomic ions which do not undergo chemical change during the reaction, treat the ions as whole units in the same way that you would treat individual atoms. 1 Cu 1 Ag 1 NO 3 Balance NO 3 by placing a coefficient of 2 in front of AgNO 3. 1 Cu 1 Ag 2 NO 3 Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + Ag(s) 1 Cu 2 Ag 2 NO 3 1 Cu 1 Ag 2 NO 3 The equation is balanced when we add a coefficient of 2 for Ag(s). Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s) 5.42 The formulas for barium and chloride ions are Ba 2+ and Cl, respectively. The formula for aqueous barium chloride is BaCl 2 (aq), and the formulas for sulfuric acid and hydrochloric acid are H 2 SO 4 (aq) and HCl(aq), respectively. Since the charges on barium and sulfate ions are of the same magnitude (2+ and 2-, respectively), the formula of barium sulfate is BaSO 4 (s) (Table 5.3). See Figures 3.12 and 3.17 and the nomenclature rules for acids in Chapter 3 if you need more help with these formulas. The unbalanced equation is: BaCl 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + HCl(aq) Unbalanced When balancing chemical equations that contain polyatomic ions which do not undergo chemical change during the reaction, treat the ions as whole units in the same way that you would treat individual atoms. 1 Ba 2 Cl 2 H 2 1 SO 4 Begin by balancing Cl. 1 Ba 1 Cl 1 H 2 1 SO 4 BaCl 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2HCl(aq) 1 Ba 2 Cl 2 H 2 1 SO 4 1 Ba 2 Cl 2 H 2 1 SO 4 Since the number of each type of atom is the same on both sides of the equation, the equation is balanced. 5.43 Table 5.1 lists the characteristics of the reactants and products for the different classifications of reactions. Decomposition: : 1 compound; : 2 elements or smaller compounds Combination: : 2 elements or compounds; : 1 compound Single-displacement: : 1 element and 1 compound; : 1 element and 1 compound Double-displacement: : 2 compounds; : 2 compounds 5.44 Table 5.1 lists the characteristics of the reactants and products for the different classifications of reactions. There are many different examples that you could have picked (if you need ideas, look through the sections of this chapter). As you select example reactions, make sure they match the descriptions given in Table 5.1. Decomposition: : 1compound; : 2 elements or smaller compounds H 2 CO 3 (aq) CO 2 (g) + H 2 O(l) Combination: : 2 elements or compounds; : 1 compound 2H 2 (g) + O 2 (g) 2H 2 O(l) 5 15

Single-displacement: : 1 element and 1 compound; : 1 element and 1 compound Cu(s) + 2AgNO 3 (aq) 2Ag(s) + Cu(NO 3 ) 2 (aq) Double-displacement: : 2 compounds; : 2 compounds CuCl 2 (aq) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2AgCl(s) 5.45 For help classifying reactions, refer to Table 5.1. We can classify two of the types based on the number of products and reactants in the chemical equation. In combination reactions, 2 substances combine to form 1 new substance; in decomposition reactions, 1 compound forms several new substances. Equations indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If one of the reactants and one of the products is an element, the reaction is single-displacement. (a) combination (b) single-displacement (c) decomposition 5.46 For help classifying reactions, refer to Table 5.1. We can classify two of the types based on the number of products and reactants in the chemical equation. In combination reactions, 2 substances combine to form 1 new substance; in decomposition reactions, 1 compound forms several new substances. Equations indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If one of the reactants and one of the products is an element, the reaction is single-displacement. (a) decomposition (b) double-displacement (c) combination 5.47 From the description we learn that there are two reactants: sodium chloride and lead(ii) nitrate. These are both compounds. There are also two products: lead(ii) chloride and sodium nitrate. Since these are also compounds, the reaction is double-displacement (2 compounds as reactants and 2 compounds as products). 5.48 The reactants are solid sulfur and oxygen gas. These are both elements. Since there is only one product, sulfur dioxide, which is a compound, this is a combination reaction (2 elements or compounds as reactants and 1compound as the product). 5.49 (a) There are two different substances in the reactant image and only one substance in the product image. This is a combination reaction. (b) The dark red and gray spheres of solid represent two different elements. The gray spheres are displacing the dark red spheres. One compound and 1 element are producing another compound and element. This is a single-displacement reaction. 5.50 (a) The image on the left represents some type of ionic compound. On the right, the different types of atoms have recombined into elements. This is a decomposition reaction. (b) The image on the left represents an aqueous solution of two different ionic compounds. The image on the right indicates the presence of two new compounds, one of which is water soluble and one that is not. This is a double displacement reaction. 5.51 For help classifying reactions, refer to Table 5.1. We can classify two of the types based on the number of products and reactants in the chemical equation. In combination reactions, 2 substances combine to form 1 new substance. In decomposition reactions, 1 compound forms several new substances. Equations indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If 1 of the reactants and 1 of the products is an element, the reaction is single-displacement. (a) double-displacement: CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s) + 2NaCl(aq) (b) single-displacement: Ba(s) + 2HCl(aq) BaCl 2 (aq) + H 2 (g) (c) combination: N 2 (g) + 3H 2 (g) 2NH 3 (g) (d) This reaction does not easily fit the classification scheme given in Table 5.1; however, it is most like a single-displacement reaction. CO is displacing iron from FeO: heat FeO(s) + CO(g) Fe(s) + CO 2 (g) 5 16

(e) combination: CaO(s) + H 2 O(l) Ca(OH) 2 (aq) (f) double-displacement: Na 2 CrO 4 (aq) + Pb(NO 3 ) 2 (aq) PbCrO 4 (s) + 2NaNO 3 (aq) (g) single-displacement: 2KI(aq) + Cl 2 (g) 2KCl(aq) + I 2 (aq) heat (h) decomposition: 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) 5.52 For help classifying reactions, refer to Table 5.1. We can classify two of the types on the number of products and reactants in the chemical equation. In combination reactions, 2 substances combine to form 1 new substance; in decomposition reactions, 1 compound forms several new substances. Equations indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If 1 of the reactants and 1 of the products is an element, the reaction is single-displacement. (a) combination: GaH 3 + N(CH 3 ) 3 (CH 3 ) 3 NGaH 3 (b) single-displacement: Ca(s) +2H 2 O(l) Ca(OH) 2 (aq) + H 2 (g) (c) single-displacement: N 2 (g) + CaC 2 (s) 2C(s) + CaNCN(s) (d) combination: N 2 (g) + 3Mg(s) Mg 3 N 2 (s) (e) decomposition: NH 4 Cl(s) NH 3 (g) + HCl(g) heat (f) combination: CaO(s) + SO 3 (g) CaSO 4 (s) heat (g) decomposition: PCl 5 (g) PCl 3 (g) + Cl 2 (g) heat (h) double-displacement: Ca 3 N 2 (s) + 6H 2 O(l) 3Ca(OH) 2 (aq) + 2NH 3 (g) 5.53 Carbonate-containing compounds (except Group IA (1)), when heated, produce the metal oxide and carbon dioxide gas (Table 5.2). In this case, we know that nickel has a 2+ charge to balance the charge of the 2 carbonate ion. The oxide will be NiO. The balanced chemical equation is NiCO 3 (s) NiO(s) + CO 2 (g). 5.54 When chloride-containing compounds are heated they decompose, forming chlorine gas and the elemental metal (Table 5.2). In this case, platinum(iv) chloride, PtCl 4, decomposes to produce platinum metal, Pt, and chlorine gas, Cl 2. The balanced chemical equation is PtCl 4 (s) Pt(s) + 2Cl 2 (g). Note: Most ionic compounds (such as PtCl 4 ) are solids, and molecular chlorine is a gas. 5.55 See Table 5.2. (a) CaCO 3 (s) heat CaO(s) + CO 2 (g) Metal carbonates (except Group IA (1)) decompose to produce carbon dioxide and the metal oxide. We determine the formula of the metal oxide from the charge on the metal and the 2 charge of the oxide ion (i.e. Ca 2+ and O 2 produce CaO). (b) CuSO 4 5H 2 O(s) 5.56 See Table 5.2. heat CuSO 4 (s) + 5H 2 O(g) Hydrates decompose by separation of the water from the ionic compound. (a) Cu(OH) 2 (s) heat CuO(s) + H 2 O(g) Hydroxides decompose to the metal oxide with the loss of water (as a vapor or gas). Even at the high temperatures needed to decompose many of these compounds, ionic compounds remain in the solid state. (b) 2NaN 3 (s) 5.57 See Table 5.3 heat 2Na(s) + 3N 2 (g) Azides often decompose to produce nitrogen gas. 2Mg(s) + O 2 (g) 2MgO(s) Combination reactions result in the formation of 1 product from 2 reactants. If the reactants are a metal and a nonmetal, we can reliably predict that the metal will form a cation and the nonmetal will form an anion. 5 17

When oxygen reacts with elements, the oxides of those elements are formed. With metals, the products are the metal oxides. Magnesium is a Group IIA (2) metal. This means that it will form an ion with a 2+ charge. The formula for oxide ion is O 2. The formula for the metal oxide product in this reaction is MgO. 5.58 See Table 5.3 2Na(s) + Cl 2 (g) 2NaCl(s) Combination reactions result in the formation of 1 product from 2 reactants. If the reactants are a metal and a nonmetal, we can reliably predict that the metal will form a cation and the nonmetal will form an anion. When molecular chlorine, Cl 2, reacts with metals, the chlorides of the metals are formed. Sodium is a Group IA (1) metal so it will form a 1+ ion. The formula for chloride ion is Cl. The formula for the metal chloride is NaCl (salt). 5.59 Combination reactions result in the formation of 1 product from 2 reactants (see Table 5.3). If the reactants are a metal and a nonmetal, you can reliably predict that the metal will form a cation and the nonmetal will form an anion. In other situations, simply try to combine the two reactants to form a product you recognize. (a) 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Calcium and nitrogen (metal and nonmetal) form the Ca 2+ and N 3 ions based on their positions on the periodic table. We predict that the product they form is Ca 3 N 2. (b) 2K(s) + Br 2 (l) 2KBr(s) Metals and nonmetals often react to form metal salts. (c) 4Al(s) +3O 2 (g) 2Al 2 O 3 (s) Aluminum and oxygen (metal and nonmetal) form Al 3+ and O 2 ions, based on their positions on the periodic table. We predict that the product they form is Al 2 O 3. 5.60 Combination reactions result in the formation of 1 product from 2 reactants (see Table 5.3). Metals and nonmetals often react to form metal salts. Each of these products is the metal salt formed the two reactants. (a) 4Na(s) + O 2 (g) 2Na 2 O(s) (b) 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) (c) Ca(s) + F 2 (g) CaF 2 (s) 5.61 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a metal displaces the metal ion, or a nonmetal displaces the nonmetal ion. To determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the metal whose ion appears in the compound, a reaction occurs. In these cases, the formula for the ionic product is determined by the metal ion that is produced. If you can t predict the charge (i.e. it is not listed in Figure 3.12), assume that it has a charge of 2+. For example if Fe displaces Cu, Fe will form a 2+ ion. In this type of single-displacement reaction, the formula for the anion remains unchanged. (a) Zn is higher on the activity series than Ag so a reaction occurs. Zn forms a 2+ ion, and Ag + is displaced, forming silver metal, Ag(s). Zn(s) + 2AgNO 3 (aq) 2Ag(s) + Zn(NO 3 ) 2 (aq) (b) Na is higher on the activity series than Fe so a reaction occurs. Na forms a 1+ ion, and Fe 2+ is displaced, forming iron metal, Fe(s). 2Na(s) + FeCl 2 (s) 2NaCl(s) + Fe(s) 5. 62 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a metal displaces the metal ion, or a nonmetal displaces the nonmetal ion. To determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the metal whose ion appears in the compound, a reaction occurs. In these cases, the formula for the ionic 5 18

product is determined by the metal ion that is produced. If you can t predict the charge (i.e. it is not listed in Figure 3.12), assume that it has a charge of 2+. For example if Fe displaces Cu, Fe will form a 2+ ion. In this type of single-displacement reaction, the formula for the anion remains unchanged. (a) Al (s) higher on the activity series than Cu so a reaction occurs. Al forms a 3+ ion, and Cu 2+ is displaced as copper metal, Cu(s). 2Al(s) + 3CuSO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 3Cu(s) (b) Cs is higher in activity than H 2 so a reaction occurs. In reactions of active metals with water, hydrogen gas is formed. Cs forms a 1+ ion, and H + is displaced as H 2 (g). When metals displace H + from H 2 O (or HOH), the product is a metal hydroxide. 2Cs(s) + 2H 2 O(l) 2CsOH(aq) + H 2 (g) 5.63 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a metal displaces the metal ion or a nonmetal displaces the nonmetal ion. The formulas for the products are determined by the element or ions that are produced (i.e. metal or diatomic molecules, etc.). When Zn displaces Sn, Zn will form a 2+ ion (Figure 3.12). In this single-displacement reaction, the formula for the anion remains unchanged. Since the formula for chloride ion is Cl, the product is ZnCl 2. The balanced equation is: Zn(s) + SnCl 2 (aq) ZnCl 2 (aq) + Sn(s) 5.64 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a metal displaces the metal ion or a nonmetal displaces the nonmetal ion. The formulas for the products are determined by the element or ions that are produced (i.e. metal or diatomic molecules, etc.). When Mg displaces Cu, Mg will form a 2+ ion. In this single-displacement reaction, the formula for the anion remains unchanged. Mg(s) + CuSO 4 (aq) Cu(s) + MgSO 4 (aq) 5.65 In reactions of metals with hydrochloric acid solution or water, one product will always be hydrogen gas (H 2 ). To determine whether a reaction takes place, compare the activity of the metal (not the metal ion found in the product) with the activity of hydrogen gas (Figure 5.21). If the metal is higher in the activity series than hydrogen gas, a reaction will take place. Some metals are more active than others. Those that are more active react under less extreme conditions. (a) Ca is more active than H 2. Because it is so high on the activity series, it will displace hydrogen from water and form an acidic solution. As noted on the activity series shown in Figure 5.21, if a substance reacts with cold water, it also reacts with steam and acid. Ca reacts with water: Ca(s) + 2H 2 O(l) Ca(OH) 2 (aq) + H 2 (g) Ca reacts with HCl: Ca(s) + 2HCl(aq) CaCl 2 (aq) + H 2 (g) (b) Fe is more active than H 2. This means that Fe can displace H 2 under the proper conditions. As is noted in Figure 5.21, Fe is not active enough to displace hydrogen in cold water, but can do so from either hydrochloric acid solution or from steam. Fe reacts with steam: Fe(s) + 2 H 2 O(g) Fe(OH) 2 (s) + H 2 (g) Fe reacts with HCl: Fe(s) + 2HCl(aq) FeCl 2 (aq) + H 2 (g) (c) No reaction. Copper is below H 2 on the activity series. There are no reaction conditions which favor a reaction of copper with water or HCl solutions. 5.66 In reactions of metals with hydrochloric acid solutions or water, one product will always be hydrogen gas (H 2 ). To determine whether a reaction takes place, compare the activity of the metal (not the metal ion found in the product) with the activity of hydrogen gas (Figure 5.21). If the metal is higher in the activity series than hydrogen gas, a reaction will take place. Some metals are more active than others. Those that are more active react under less extreme conditions. 5 19

(a) Cr is more active than H 2. This means that it can displace hydrogen under the right conditions. However, Cr is not active enough to displace hydrogen in cold water. Either hydrochloric acid solution or steam is required to cause a reaction. Chromium forms a 2+ ion in the activity series. Cr reacts with HCl: Cr(s) + 2HCl(aq) CrCl 2 (aq) + H 2 (g) Cr reacts with steam: Cr(s) + 2H 2 O(g) Cr(OH) 2 (s) + H 2 (g) (b) No reaction. Bismuth is below H 2 on the activity series. There are no reaction conditions which favor a reaction of bismuth with water or HCl solutions. (c) K is more active than H 2. Because it is so high on the activity series, K will displace hydrogen from water and from hydrochloric acid solution. As noted on the activity series, if a substance reacts with cold water, it also reacts with steam and acid. K reacts with water: 2K(s) + 2H 2 O(l) 2KOH(aq) + H 2 (g) K reacts with HCl: 2K(s) + 2HCl(aq) 2KCl(aq) + H 2 (g) 5.67 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a metal displaces the metal ion, or a nonmetal displaces the nonmetal ion. To determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the metal whose ion appears in the compound, a reaction occurs. (a) Magnesium is more active than aluminum. The reaction occurs. (b) Zinc is less active than magnesium. No reaction occurs. (c) Copper is less active than lead. No reaction occurs. (d) Nickel is more active than silver. The reaction occurs. 5.68 In a single displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a metal displaces the metal ion, or a nonmetal displaces the nonmetal ion. To determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the metal whose ion appears in the compound, a reaction occurs. (a) Nickel is less active than iron. No reaction occurs. (b) Aluminum is more active than nickel. The reaction occurs. (c) Iron is more active than copper. The reaction occurs. (d) Tin is less active than aluminum. No reaction occurs. 5.69 To determine solubility, refer to the solubility rules given in Table 5.4. Except for compounds containing NH 4 + and/or Group IA (1) cations (including Na + and K + ), the solubility rules focus on the anions in ionic compounds. It is worth memorizing that Group IA (1), nitrates, acetates, and ammonium ion compounds are soluble, because we encounter them so frequently when we study chemistry. (a) CuCl 2 : The anion is Cl ; the compound is soluble. (b) AgNO 3 : The anion is NO 3 ; the compound is soluble. (c) PbCl 2 : The anion is Cl ; most chlorides are soluble, but PbCl 2 is an exception. It is insoluble. (d) Cu(OH) 2 : The anion is OH ; the compound is insoluble. 5.70 To determine solubility, refer to the solubility rules given in Table 5.4. Except for compounds containing NH 4 + and/or Group IA (1) cations (including Na + and K + ), the solubility rules focus on the anions in ionic compounds. It is worth memorizing that Group IA (1), nitrates, acetates, and ammonium ion compounds are soluble, because we encounter them so frequently when we study chemistry. (a) Cr 2 S 3 : The anion is S 2 ; the compound is insoluble. (b) Ca(OH) 2 : The anion is OH ; the compound is soluble. The solubility rules indicate that Ca(OH) 2 is somewhat soluble. This means that a small amount of Ca(OH) 2 will dissolve in water. (c) BaSO 4 : The anion is SO 4 2 ; the compound is insoluble. While most sulfates are soluble in water, there are exceptions, such as BaSO 4. (d) (NH 4 ) 2 CO 3 : The cation is NH 4 + ; the compound is soluble. All ammonium compounds are soluble. 5 20

5.71 For each reaction we: 1) determine the ion formulas, 2) predict the products based on ionic charges, and 3) determine the states of the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. Finally, we balance the resulting equation. (a) : K 2 CO 3 (aq) and BaCl 2 (aq) Ions: K +, CO 3 2, Ba 2+, Cl : BaCO 3 and KCl Solubility (from Table 5.4): BaCO 3 (s) and KCl(aq) equation: K 2 CO 3 (aq) + BaCl 2 (aq) BaCO 3 (s) + 2KCl(aq) (b) : CaS(aq) and Hg(NO 3 ) 2 (aq) Ions: Ca 2+, S 2, Hg 2+, NO 3 : Ca(NO 3 ) 2 and HgS Solubility (from Table 5.4): Ca(NO 3 ) 2 (aq) and HgS(s) equation: CaS(aq) + Hg(NO 3 ) 2 (aq) Ca(NO 3 ) 2 (aq) + HgS(s) (c) : Pb(NO 3 ) 2 (aq) and K 2 SO 4 (aq) Ions: Pb 2+, NO 3, K +, SO 4 2 : PbSO 4 and KNO 3 Solubility (from Table 5.4): PbSO 4 (s) and KNO 3 (aq) equation: Pb(NO 3 ) 2 (aq) + K 2 SO 4 (aq) PbSO 4 (s) + 2KNO 3 (aq) 5.72 For each reaction we: 1) determine the ion formulas, 2) predict the products based on ionic charges, and 3) determine the states of the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. Finally, we balance the resulting equation. (a) : MgSO 4 (aq) and BaCl 2 (aq) Ions: Mg 2+, SO 4 2, Ba 2+, Cl : MgCl 2 and BaSO 4 Solubility (from Table 5.4): MgCl 2 (aq) and BaSO 4 (s) equation: MgSO 4 (aq) + BaCl 2 (aq) MgCl 2 (aq) and BaSO 4 (s) (b) : K 2 SO 4 (aq) and MgCl 2 (aq) Ions: K +, SO 4 2, Mg 2+, Cl : KCl and MgSO 4 Solubility (from Table 5.4): KCl(aq) and MgSO 4 (aq) equation: No reaction occurs because both products are ionic and soluble in water. (c) : MgCl 2 (aq) and Pb(NO 3 ) 2 (aq) Ions: Mg 2+, Cl, Pb 2+, NO 3 : Mg(NO 3 ) 2 and PbCl 2 Solubility (from Table 5.4): Mg(NO 3 ) 2 (aq) and PbCl 2 (s) equation: MgCl 2 (aq) + Pb(NO 3 ) 2 (aq) Mg(NO 3 ) 2 (aq) + PbCl 2 (s) 5.73 For each reaction, 1) determine the ion formulas, 2) predict the products based on ion charges, and 3) determine the states of the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. Balance the resulting equations. (a) : BaCO 3 (s) and H 2 SO 4 (aq) Ions: Ba 2+, CO 3 2, H +, SO 4 2 : BaSO 4 and H 2 CO 3 According to the solubility rules (Table 5.4), barium sulfate is insoluble in water. Carbonic acid decomposes to carbon dioxide and water (as described in Table 5.2). Since a precipitate is formed and carbonic acid decomposes to form water and carbon dioxide, we predict that the following reaction takes place: BaCO 3 (s) + H 2 SO 4 (aq) BaSO 4 (s) + H 2 CO 3 (aq) carbonic acid decomposes 5 21

BaCO 3 (s) + H 2 SO 4 (aq) BaSO 4 (s) + H 2 O(l) + CO 2 (g) (b) : CuCl 2 (aq) and AgNO 3 (aq) Ions: Cu 2+ (based on charge of chloride), Cl, Ag +, NO 3 : Cu(NO 3 ) 2 and AgCl According to the solubility rules (Table 5.4), nitrates are soluble in water. While most chlorides are soluble in water, AgCl is an exception. Since a precipitate is formed, we can predict that the following reaction takes place: CuCl 2 (aq) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2AgCl(s) 5.74 For each reaction, 1) determine the ion formulas, 2) predict the products based on ion charges, and 3) determine the states of the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. Balance the resulting equations. (a) : NaOH(aq) and CuCl 2 (aq) Ions: Na +, OH, Cu 2+ (based on the charge of chloride ion), Cl : NaCl and Cu(OH) 2 According to the solubility rules (Table 5.4), most chlorides are soluble in water, so NaCl will be dissolved. Most hydroxides are insoluble in water, so Cu(OH) 2 will form a precipitate (that is, a solid). Because the precipitate forms, we predict that the following reaction takes place: 2NaOH(aq) + CuCl 2 (aq) 2NaCl(aq) + Cu(OH) 2 (s) (b) : H 2 SO 4 (aq) and KOH(aq) Ions: H +, SO 4 2, K +, OH : H 2 O and K 2 SO 4 According to the solubility rules (Table 5.4), sulfates are generally insoluble in water; however, K 2 SO 4 is an exception. Because water is a stable molecule, we can predict that the following reaction takes place: H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) 5.75 To identify the precipitate, we first determine the products of the reaction, and then check the solubility of the products using the solubility rules (Table 5.4). The precipitate will be the compound that is insoluble in water. We deduce the reactants from the description given in the problem. : Na 2 SO 4 and Pb(NO 3 ) 2 Ions: Na +, SO 4 2, Pb 2+, NO 3 : PbSO 4 and NaNO 3 Solubility (from Table 5.4): PbSO 4 (s), NaNO 3 (aq) Lead(II) sulfate is the white solid that forms when the two solutions are mixed. 5.76 To identify the precipitate, we first determine the products of the reaction, and then check the solubility of the products using the solubility rules (Table 5.4). The precipitate will be the compound that is insoluble in water. We deduce the reactants from the description given in the problem. In this particular problem, we encounter the mercury(i) ion. Mercury(I) ions (Hg 2 2+ ) represent one of a very few diatomic metal ions (see Figure 3.25). Notice that the ion has a 2+ charge because it is diatomic. The average charge of mercury in this ion is 1+. : Hg 2 (NO 3 ) 2 and CaI 2 Ions: Hg 2 2+, NO 3, Ca 2+, I : Hg 2 I 2 and Ca(NO 3 ) 2 Solubility (from Table 5.3): Hg 2 I 2 (s), Ca(NO 3 ) 2 (aq) Mercury(I) iodide is the yellow solid that forms when the two solutions are mixed. 5.77 From the problem description we can write: CaCl 2 (aq) + K 2 CO 3 (aq) 5 22

In double-displacement reactions, the cations switch places to form new products (i.e. the metal ions displace one another). To avoid common errors, and to be able to spot mistakes easily, it is helpful if we write the complete formulas for each ion before we determine the products. : CaCl 2 (aq) and K 2 CO 3 (aq) Ions: Ca 2+, Cl, K +, CO 3 2 Considering the charges of the ions, we can write the chemical formulas for the products. The products will be calcium carbonate, CaCO 3, and potassium chloride, KCl. Using the solubility rules (Table 5.4) we can determine whether these products are soluble or insoluble in water (i.e. whether we should write (aq), or (s), respectively). From the solubility rules, we determine that carbonates are insoluble (i.e. CaCO 3 (s)) and chlorides are generally soluble (i.e. KCl(aq)). CaCl 2 (aq) + K 2 CO 3 (aq) CaCO 3 (s) + 2KCl(aq) 5.78 From the problem description we can write: (NH 4 ) 2 CrO 4 (aq) + Pb(NO 3 ) 2 (aq) In double-displacement reactions, the cations switch places to form new products (i.e. the metal ions displace one another). To avoid common errors, and to be able to spot mistakes easily, it is helpful if we write the complete formulas for each ion before we determine the products. : (NH 4 ) 2 CrO 4 (aq) and Pb(NO 3 ) 2 (aq) Ions: NH 4 +, CrO 4 2, Pb 2+, NO 3 Considering the charges of the ions, we can write the chemical formulas for the products. The products will be lead(ii) chromate, PbCrO 4, and ammonium nitrate, NH 4 NO 3. Using the solubility rules (Table 5.4) we can determine whether these products are soluble or insoluble in water (i.e. whether we should write (aq) or (s), respectively). From the solubility rules we determine that chromates are generally insoluble in water and compounds containing ammonium and/or nitrate ions are generally soluble in water. (NH 4 ) 2 CrO 4 (aq) + Pb(NO 3 ) 2 (aq) PbCrO 4 (s) + 2NH 4 NO 3 (aq) 5.79 Chemical reactions occur for a reason. For many reactions, the reason is the nature of the products that form. This driving force of a reaction can often be identified by formation of a precipitate, formation of a gas (water-insoluble gas), or formation of molecular compounds, such as water. Water-insoluble gases will leave a solution, providing an additional driving force for the reaction. The driving force for each reaction is: (a) precipitation of HgS(s) (b) formation of the insoluble gas H 2 S(g) (c) formation of the molecular compound H 2 O(l), and precipitation of BaSO 4 (s) 5.80 Chemical reactions occur for a reason. For many reactions, the reason is the nature of the products that form. This driving force of a reaction can often be identified by formation of a precipitate, formation of a gas (water-insoluble gas), or formation of molecular compounds, such as water. Water-insoluble gases will leave a solution, providing an addition driving force for the reaction. The driving force for each reaction is: (a) formation of the molecular compound H 2 O(l) (b) formation of a precipitate, CaSO 4 (s), a molecular compound, H 2 O(l), and a gas, CO 2 (g) (c) formation of a precipitate, BaSO 4 (s) 5.81 Neutralization reactions involve the reaction of an acid with a base. The products are a salt (i.e. ionic compound) and water. Just as with precipitation reactions we: 1) determine the ionic formulas, 2) predict the products based on ion charges, and 3) determine the states of the products. Acid-base reactions are usually driven by the formation of water, a molecular compound. We must use the solubility rules to determine whether or not the salt will dissolve in water. (a) : H 2 S(aq) and Cu(OH) 2 (s) Ions: H +, S 2, Cu 2+, OH : H 2 O(l) and CuS(s) (determined from solubility rules) equation: H 2 S(aq) + Cu(OH) 2 (s) CuS(s) + 2H 2 O(l) 5 23

(b) CH 4 is not an acid, so no reaction is expected. (c) : KHSO 4 (aq) and KOH(aq) Ions: K +, H +, SO 4 2, OH : H 2 O(l) and K 2 SO 4. Note that the H + reacts with OH, leaving K + and SO 4 2 to form the salt, which is water-soluble. equation: KHSO 4 (aq) + KOH(aq) K 2 SO 4 (aq) + H 2 O(l) 5.82 Neutralization reactions involve the reaction of an acid with a base. The products are a salt (i.e. ionic compound) and water. Just as with precipitation reactions we: 1) determine the ion formulas, 2) predict the products based on ion charges, and 3) determine the states of the products. Acid-base reactions are usually driven by the formation of water (a molecular compound). We must use the solubility rules to determine whether or not the salt will dissolve in water. (a) H 2 (g) is not an acid, so no reaction is expected. (b) : H 3 PO 4 (aq) and NaOH(aq) Ions: H +, PO 4 3-, Na +, OH : Na 3 PO 4 (aq) and H 2 O(l) equation: H 3 PO 4 (aq) + 3NaOH(aq) Na 3 PO 4 (aq) + 3H 2 O(l) (c) : H 2 SO 4 (aq) and Fe(OH) 2 (aq) Ions: H +, SO 4 2, Fe 2+, OH : FeSO 4 (aq) and H 2 O(l) equation: H 2 SO 4 (aq) + Fe(OH) 2 (aq) FeSO 4 (aq) + 2H 2 O(l) 5.83 Oxygen, O 2 (g) is always one of the reactants of a combustion reaction. 5.84 Carbon dioxide and water are the usual products of hydrocarbon combustion. 5.85 The combustion reaction products are the oxides of those elements. For metals, we can predict the oxide formula based on the common charges of the metal ions and the charge of oxide ion, O 2. Metals that form several different cations can form several different oxides. (a) Cesium only forms a 1+ ion, so the oxide that forms is Cs 2 O(s). (b) Lead forms either a 2+ or 4+ ion (Figure 3.24), so the oxides that lead can form are PbO(s) and PbO 2 (s). (c) Aluminum only forms a 3+ ion, so the formula of aluminum oxide is Al 2 O 3 (s). (d) Combustion of hydrogen produces H 2 O(g). (e) Carbon produces two different oxides, CO(g) and CO 2 (g). 5.86 The combustion reaction products are the oxides of those elements. For metals, we can predict the oxide formula based on the common charges of the metal ions and the charge of oxide ion, O 2. Metals that form several different cations (see Table 3.4) can form several different oxides. (a) Magnesium only forms a 2+ ion, so the oxide that forms is MgO(s). (b) Iron forms either a 2+ or 3+ ion, so the oxides that iron can form are FeO(s) and Fe 2 O 3 (s). (c) Lithium only forms a 1+ ion, so the formula of lithium oxide is Li 2 O(s). (d) Combustion of S 8 produces SO 2 (g). (e) Nitrogen produces many different oxides. Some examples are NO and NO 2. 5.87 The combustion reaction products of compounds are the oxides of the elements which make up the compound. In many cases, there are multiple products. (a) CH 4 (and all hydrocarbons) combusts to form oxides of carbon and hydrogen. The products are CO(g) or CO 2 (g) and H 2 O(g). (b) The combustion of CO results in formation of CO 2 (g). (c) The combustion of aluminum metal results in formation of solid aluminum oxide solid, Al 2 O 3 (s). (d) CH 3 OH (and all compounds containing C, H, and O) combusts to form CO(g) or CO 2 (g) and H 2 O(g). 5 24

5.88 The combustion reaction products of compounds are the oxides of the elements which make up the compound. In many cases, there are multiple products. (a) The combustion of sodium metal results in the formation of sodium oxide, Na 2 O(s). (b) The combustion of elemental phosphorus results in the formation of diphosphorus pentoxide solid, P 2 O 5 (s). (c) C 3 H 8 (and all hydrocarbons) combusts to form oxides of carbon and hydrogen. The products are CO(g) or CO 2 (g) and H 2 O(g). (d) HCO 2 H (and all compounds containing C, H, and O) combust to form the oxides of carbon and hydrogen. The products are CO(g) or CO 2 (g) and H 2 O(g). 5.89 An electrolyte produces ions when it dissolves in water. A nonelectrolyte does not produce ions when it dissolves in water. 5.90 Dissolve the solid, then determine whether or not the resulting liquid conducts electricity with an apparatus such as that depicted in Figure 3.3. If the solid is not soluble in water, melt it and then see if it conducts electricity. 5.91 Aqueous solutions of soluble ionic compounds, acids, or bases are electrolytes because the substances produce ions when they dissolve. (a) NaOH(aq): base; electrolyte (b) HCl(aq): acid; electrolyte (c) C 12 H 22 O 11 : not a soluble ionic compound, acid, or base; nonelectrolyte 5.92 Aqueous solutions of soluble ionic compounds, acids, or bases are electrolytes because the substances produce ions when they dissolve. (a) CH 3 CH 2 OH: not a soluble ionic compound, acid, or base; nonelectrolyte (b) H 2 O(l): not a soluble ionic compound, acid, or base; nonelectrolyte (c) NaCl(aq): soluble ionic compound: electrolyte 5.93 The substance is a nonelectrolyte because it does not appear to have dissociated in the solution. In order to be an electrolyte, ions must form in the solution. 5.94 Since we started with a single substance and now it has formed multiple species, the substance is probably an electrolyte that has dissociated into its ions. 5.95 Soluble ionic compounds, acids, and bases produce ions when they dissolve in water. (a) C 6 H 12 O 6 : not a soluble ionic compound, acid, or base; does not form ions in solution (b) CH 4 : not a soluble ionic compound, acid, or base; does not form ions in solution (c) NaCl: soluble ionic compound; forms Na + and Cl ions in solution; Na + (aq) and Cl (aq) 5.96 Soluble ionic compounds, acids, or bases produce ions when they dissolve in water. (a) I 2 : not a soluble ionic compound, acid, or base; does not form ions in solution (b) KI: soluble ionic compound; forms K + and I ions in solution (a) CH 3 OH: not a soluble ionic compound, acid, or base; does not form ions in solution 5.97 Molecular equations: We write the chemical formulas for all substances in their complete form (for example, O 2 (g), NaCl(aq), Ag(s)). Ionic Equations: We write the formulas of all soluble salts, strong acids and bases as individual, solvated ions (for example, NaCl(aq) becomes Na + (aq) and Cl (aq)). Net ionic equations: We remove all spectator ions from the ionic equation. 5.98 It is necessary to identify substances as electrolytes or nonelectrolytes because electrolytes are written as separated ions in a net ionic equation. This allows us to identify and eliminate the spectator ions. 5 25

5.99 Spectator ions are those ions that do not actually participate in the chemical change that occurs during the reaction. These are the ions that appear both as reactants and as products in an ionic equation. 5.100 We find spectator ions only in ionic equations. We do not write compounds in ionic form in molecular equations. When we write net ionic equations, we do not include any spectator ions. 5.101 (a) From the solubility rules (Table 5.4) we find that all the substances are soluble ionic compounds except Ag 2 SO 4 and AgCl, which are insoluble in water. 2NaCl(aq) + Ag 2 SO 4 (s) Na 2 SO 4 (aq) + 2AgCl(s) Molecular equation Then we write the formulas of all soluble ionic compounds as ions, and eliminate the spectator ions. 2Na + (aq) + 2Cl (aq) + Ag 2 SO 4 (s) 2Na + (aq) + SO 2+ 4 (aq) + 2AgCl(s) 2Na + (aq) + 2Cl (aq) + Ag 2 SO 4 (s) 2Na + (aq) + SO 2+ 4 (aq) + 2AgCl(s) This leaves the balanced net ionic equation. 2Cl (aq) + Ag 2 SO 4 (s) 2AgCl(s) +SO 4 2 (aq) Net ionic equation Ionic equation (b) From the solubility rules we know that Cu(OH) 2 is insoluble, while the other ionic compounds are soluble. Water is a liquid. Cu(OH) 2 (s) + 2HCl(aq) CuCl 2 (aq) + 2H 2 O(l) Molecular equation Next we write the ionic equation by writing all the aqueous ionic compounds and strong acids (Table 3.10) in ionic form. Cu(OH) 2 (s) + 2H + (aq) + 2Cl (aq) Cu 2+ + 2Cl (aq) + 2H 2 O(l) Finally, we write the net ionic equation by identifying and eliminating the spectator ions. Cu(OH) 2 (s) + 2H + (aq) + 2Cl (aq) Cu 2+ + 2Cl (aq) + 2H 2 O(l) Cu(OH) 2 (s) + 2H + (aq) Cu 2+ (aq) + 2H 2 O(l) Net ionic equation (c) From the solubility rules we know that only BaCl 2 is soluble, while the other ionic compounds are insoluble. BaCl 2 (aq) + Ag 2 SO 4 (s) BaSO 4 (s) + 2AgCl(s) Molecular equation Only BaCl 2 forms ions in solution. Since none of the products form ions, there are no spectator ions. Ba 2+ (aq) + 2Cl (aq) + Ag 2 SO 4 (s) BaSO 4 (s) + 2AgCl(s) Net ionic equation 5.102 (a) From the solubility rules (Table 5.4) we find that all the substances are soluble ionic compounds except Cr(OH) 3, which is insoluble. Cr 2 (SO 4 ) 3 (aq) + 6KOH(aq) 2Cr(OH) 3 (s) + 3K 2 SO 4 (aq) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds into their respective ions. 2Cr 3+ (aq) + 3SO 4 2 (aq) + 6K + (aq) + 6OH (aq) 2Cr(OH) 3 (s) + 6K + (aq) + 3SO 4 2 (aq) Finally, we write the net ionic equation by identifying and eliminating the spectator ions, and reducing the coefficients as much as possible. 2Cr 3+ (aq) + 3SO 4 2 (aq) + 6K + (aq) + 6OH (aq) 2Cr(OH) 3 (s) + 6K + (aq) + 3SO 4 2 (aq) 2Cr 3+ (aq) + 6OH (aq) 2Cr(OH) 3 (s) Cr 3+ (aq) + 3OH (aq) Cr(OH) 3 (s) Net ionic equation (b) From the solubility rules we find that all the substances are soluble ionic compounds except PbCl 2 (s) and PbCrO 4, which are insoluble. 5 26

K 2 CrO 4 (aq) + PbCl 2 (s) 2KCl(aq) + PbCrO 4 (s) Molecular equation Then we write the formulas of all soluble ionic compounds as ions, and eliminate the spectator ions. 2K + (aq) + CrO 4 2 (aq) + PbCl 2 (s) 2K + (aq) + 2Cl (aq) + PbCrO 4 (s) 2K + (aq) + CrO 4 2 (aq) + PbCl 2 (s) 2K + (aq) + 2Cl (aq) + PbCrO 4 (s) CrO 4 2 (aq) + PbCl 2 (s) 2Cl (aq) + PbCrO 4 (s) Net ionic equation (c) From the solubility rules we find that Na 2 SO 3 and Na 2 SO 4 are soluble ionic compounds. Sulfuric acid, H 2 SO 4, is a strong acid and SO 2 is a gas. Na 2 SO 3 (aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + H 2 O(l) + SO 2 (g) Molecular equation Next we write the ionic equation by writing all the aqueous ionic compounds and strong acids as separate ions. Although sulfuric acid is a strong acid, only the first hydrogen atom ionizes completely. 2Na + (aq) + SO 3 2 (aq) + 2H + (aq) + SO 4 (aq) 2Na + + SO 4 2 (aq) + H 2 O(l) + SO 2 (g) Finally, we write the net ionic equation by identifying and eliminating the spectator ions. 2Na + (aq) + SO 3 2 (aq) + 2H + (aq) + SO 4 2 (aq) 2Na + + SO 4 2 (aq) + H 2 O(l) + SO 2 (g) SO 3 2 (aq) + 2H + (aq) H 2 O(l) + SO 2 (g) Net ionic equation 5.103 From the solution descriptions, we know that there are four ions present in the solution before any precipitation takes place: Na +, CO 3 2, Ca 2+, and Br. (a) The only possible precipitate is CaCO 3 (see Table 5.4). The other combinations represent soluble compounds. (b) Na 2 CO 3 (aq) + CaBr 2 (aq) CaCO 3 (s) + 2NaBr(aq) (c) 2Na + (aq) + CO 3 2 (aq) + Ca 2+ (aq) + 2Br (aq) CaCO 3 (s) + 2Na + (aq) + 2Br (aq) (d) Na + (aq) and Br (aq) (e) CO 3 2 (aq) + Ca 2+ (aq) CaCO 3 (s) 5.104 From the solution descriptions, we know that there are four ions present in the solution before any precipitation takes place: K +, CrO 4 2, Ba 2+, and Cl. (a) Although most chromates are insoluble, potassium chromate is an exception. The only possible precipitate is barium chromate, BaCrO 4 (see Table 5.4). (b) K 2 CrO 4 (aq) + BaCl 2 (aq) BaCrO 4 (s) + 2KCl(aq) (c) 2K + (aq) + CrO 4 2 (aq) + Ba 2+ (aq) + 2Cl (aq) BaCrO 4 (s) + 2K + (aq) + 2Cl (aq) (d) K + (aq) and Cl (aq) (e) CrO 4 2 (aq) + Ba 2+ (aq) BaCrO 4 (s) 5.105 To write the net ionic equation for this reaction, we start by writing the molecular equation. From the description in the question we have: CaCl 2 (aq) + 2AgNO 3 (aq) Ca(NO 3 ) 2 (aq) + 2AgCl(s) Molecular equation Next we write the ionic equation by separating the formulas of all the water-soluble ionic compounds into their respective ions. All the ions written in the equation below are solvated by water. Ca 2+ (aq) + 2Cl (aq) + 2Ag + (aq) + 2NO 3 (aq) Ca 2+ (aq) + 2NO 3 (aq) + 2AgCl(s) Next we eliminate the spectator ions. Ca 2+ (aq) + 2Cl (aq) + 2Ag + (aq) + 2NO 3 (aq) Ca 2+ (aq) + 2NO 3 (aq) + 2AgCl(s) 2Cl (aq) + 2Ag + (aq) 2AgCl(s) Because each of the coefficients is divisible by two, we simplify the net ionic equation to: Ag + (aq) + Cl (aq) AgCl(s) Net ionic equation 5 27

A quick way to check the answer is to see if the total charge on the reactant and product sides of the equation is the same. In this case, the total charge is zero for both the reactants (Ag + and Cl ) and the products. 5.106 To write the net ionic equation for this reaction, we start by writing the molecular equation. From the description in the question we have: Cr 2 (SO 4 ) 3 (aq) + 6NaOH(aq) 2Cr(OH) 3 (s) + 3Na 2 SO 4 (aq) Molecular equation Next we write the ionic equation by separating the formulas of all water-soluble compounds into their respective ions. All the ions written in the equation below are solvated by water. 2Cr 3+ (aq) + 3SO 4 2 (aq) + 6Na + (aq) + 6OH (aq) 2Cr(OH) 3 (s) + 6Na + (aq) + 3SO 4 2 (aq) Next we eliminate the spectator ions. 2Cr 3+ (aq) + 3SO 4 2 (aq) + 6Na + (aq) + 6OH (aq) 2Cr(OH) 3 (s) + 6Na + (aq) + 3SO 4 2 (aq) 2Cr 3+ (aq) + 6OH (aq) 2Cr(OH) 3 (s) Because each of the coefficients in this equation is divisible by 2, we simplify the net ionic equation to: Cr 3+ (aq) + 3OH (aq) Cr(OH) 3 (s) Net ionic equation A quick way to check the answer is to see if the total charge on the reactant and product sides of the equation is the same. In this case, the total charge is zero for both the reactants (the Cr 3+ is balanced by the three negative charges of OH ) and the products. 5.107 The reactants are sodium metal (you can tell from its well ordered structure) and H 2 O(l). The products are Na +, OH, and H 2. Na + and OH are ions dissolved in water. Molecular hydrogen is a gas. The balanced ionic equation is: 2Na(s) + 2H 2 O(l) 2Na + (aq) + 2OH (aq) + H 2 (g) There are no spectator ions, so the balanced net ionic equation and the balanced ionic equations are identical. 2Na(s) + 2H 2 O(l) 2Na + (aq) + 2OH (aq) + H 2 (g) Net ionic equation 5.108 Copper and aqueous silver nitrate participate in a single displacement reaction because copper is more active than silver (Figure 5.21). We know that copper nitrate dissolves in water from the diagram, because the ions are separated. Cu(s) + 2AgNO 3 (aq) 2Ag(s) + Cu(NO 3 ) 2 (aq) Molecular equation Next we write the ionic equation by showing all the soluble ionic compounds in their ionic forms. Cu(s) + 2Ag + (aq) + 2NO 3 (aq) 2Ag(s) + Cu 2+ (aq) + 2NO 3 (aq) Eliminating spectator ions we have: Cu(s) + 2Ag + (aq) + 2NO 3 (aq) 2Ag(s) + Cu 2+ (aq) + 2NO 3 (aq) Cu(s) + 2Ag + (aq) 2Ag(s) + Cu 2+ (aq) Net ionic equation A quick way to check the answer is to see if the total charge on the reactant and product sides of the equation is the same. In this case, the total charge is 2+ on both the reactant and product sides. 5.109 (a) First we write the molecular equation and determine the solubility of ionic reactants and products (Table 5.4). Cu(s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s) Molecular equation Next we write the ionic equation by writing all the aqueous ionic compounds in ionic form. Cu(s) + 2Ag + (aq) + 2NO 3 (aq) Cu 2+ (aq) + 2NO 3 (aq) + 2Ag(s) 5 28

Finally, we eliminate the spectator ions and write the net ionic equation. Cu(s) + 2Ag + (aq) + 2NO 3 (aq) Cu 2+ (aq) + 2NO 3 (aq) + 2Ag(s) Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) Net ionic equation (b) First we write the molecular equation and determine the solubility of ionic reactants and products (Table 5.4). FeO(aq) + 2HCl(aq) FeCl 2 (aq) + H 2 O(l) Molecular equation Next we write the ionic equation by writing all the aqueous ionic compounds and strong acids (Table 3.10) in ionic form. Fe 2+ (aq) + O 2 (aq) + 2H + (aq) + Cl (aq) Fe 2+ (aq) + 2Cl (aq) + H 2 O(l) Finally, we eliminate the spectator ions and write the net ionic equation. Fe 2+ (aq) + O 2 (aq) + 2H + (aq) + 2Cl (aq) Fe 2+ (aq) + 2Cl - (aq) + H 2 O(l) O 2 (aq) + 2H + (aq) H 2 O(l) Net ionic equation 5.110 (a) First we write the molecular equation and determine the solubility of ionic reactants and products (Table 5.4). Ca(s) + 2H 2 O(l) Ca(OH) 2 (aq) + H 2 (g) Molecular equation Next we write the ionic equation by writing all the aqueous ionic compounds in ionic form. Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH (aq) + H 2 (g) There are no spectator ions, so the balanced net ionic equation and the balanced ionic equations are identical. Ca(s) + 2H 2 O(l) Ca 2+ (aq) + 2OH (aq) + H 2 (g) Net ionic equation (b) First we write the molecular equation and determine the solubility of ionic reactants and products (Table 5.4). BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) Molecular equation Next we write the ionic equation by writing all the aqueous ionic compounds in ionic form. Ba 2+ (aq) + 2Cl (aq) + 2Na + (aq) + SO 4 2 (aq) BaSO 4 (s) + 2Na + (aq) + 2Cl (aq) Finally, we eliminate the spectator ions and write the net ionic equation. Ba 2+ (aq) + 2Cl (aq) + 2Na + (aq) + SO 2 4 (aq) BaSO 4 (s) + 2Na + (aq) + 2Cl - (aq) Ba 2+ (aq) + SO 2 4 (aq) BaSO 4 (s) Net ionic equation Ionic equation 5.111 For each reaction we: 1) determine the ion formulas, 2) predict the products based on ionic charges, and 3) determine the states for the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. (a) : Sr(NO 3 ) 2 (aq) and H 2 SO 4 (aq) Ions: Sr 2+, NO 3, H + 2, SO 4 : SrSO 4 and HNO 3 Solubility (Table 5.4, Table 3.10): Sr(NO 3 ) 2 (aq), H 2 SO 4 (aq), SrSO 4 (s), HNO 3 (aq) Sr(NO 3 ) 2 (aq) + H 2 SO 4 (aq) SrSO 4 (s) + 2HNO 3 (aq) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds and strong acids into their respective ions. Sr 2+ + 2NO 3 (aq) + 2H + (aq) + SO 4 2 (aq) SrSO 4 (s) + 2H + (aq) + 2NO 3 (aq) Finally, we eliminate the spectator ions: 5 29

Sr 2+ + 2NO 3 (aq) + 2H + (aq) + SO 4 2 (aq) SrSO 4 (s) + 2H + (aq) + 2NO 3 (aq) Sr 2+ (aq) + SO 4 2 (aq) SrSO 4 (s) Net ionic equation (b) : Zn(NO 3 ) 2 and Na 2 SO 4 Ions: Zn 2+, NO 3, Na +, SO 4 2 : ZnSO 4 and NaNO 3 Solubility: Both products are aqueous ionic compounds. The products do not precipitate. In addition, no molecular products are formed. No reaction takes place. (c) : CuSO 4 (aq) and BaS(s) Ions: Cu 2+, SO 4 2, Ba 2+, S 2 : CuS and BaSO 4 Solubility: CuSO 4 (aq), BaS(s), CuS(s), BaSO 4 (s) CuSO 4 (aq) + BaS(s) CuS(s) + BaSO 4 (s) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds into their respective ions. Cu 2+ (aq) + SO 4 2 (aq) + BaS(s) CuS(s) + BaSO 4 (s) Ionic & net ionic equation Since only one reactant is soluble, there are no spectator ions, so the ionic and net ionic equations are the same. (d) : NaHCO 3 and HCH 3 CO 2 Ions: Na +, HCO 3, H +, and CH 3 CO 2 (see note below) : NaCH 3 CO 2 and H 2 CO 3 (decomposes to H 2 O(l) and CO 2 (g) Solubility: NaHCO 3 (aq), HCH 3 CO 2 (aq), NaCH 3 CO 2 (aq): Acids are generally soluble. Other solubilities are obtained from Table 5.4. A reaction takes place because of the formation of water and carbon dioxide. NaHCO 3 (aq) + CH 3 CO 2 H (aq) NaCH 3 CO 2 (aq) + CO 2 (g) + H 2 O(l) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds into their respective ions. Note that HCH 3 CO 2 is a weak acid and does not completely ionize, however when it is ionized, H + and CH 3 CO 2 are formed and we use this information to predict what products are formed. Na + (aq) + HCO 3 (aq) + HCH 3 CO 2 (aq) Na + (aq) + CH 3 CO 2 (aq) + CO 2 (g) + H 2 O(l) Finally, we eliminate the spectator ions. Na + (aq) + HCO 3 (aq) + HCH 3 CO 2 (aq) Na + (aq) + CH 3 CO 2 (aq) + CO 2 (g) + H 2 O(l) HCO 3 (aq) + CH 3 CO 2 H (aq) CH 3 CO 2 (aq) + CO 2 (g) + H 2 O(l) Net Ionic Equation 5.112 For each reaction, 1) determine the ion formulas, 2) predict the products based on ion charges, and 3) determine the states for the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. (a) : (NH 4 ) 2 CO 3 and CrCl 3 Ions: NH + 4, CO 2 3, Cr 3+, Cl : NH 4 Cl and Cr 2 (CO 3 ) 3 Solubility: (NH 4 ) 2 CO 3 (aq), CrCl 3 (aq), NH 4 Cl(aq), Cr 2 (CO 3 ) 3 (s) 3(NH 4 ) 2 CO 3 (aq) + 2CrCl 3 (aq) 6NH 4 Cl(aq) + Cr 2 (CO 3 ) 3 (s) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds into their respective ions. 6NH 4 + (aq) + 3CO 3 2 (aq) + 2Cr 3+ (aq) + 6Cl (aq) 6NH 4 + (aq) + 6Cl (aq) + Cr 2 (CO 3 ) 3 (s) Finally, we write the net ionic equation by identifying and eliminating the spectator ions. 5 30

6NH 4 + (aq) + 3CO 3 2 (aq) + 2Cr 3+ (aq) + 6Cl (aq) 6NH 4 + (aq) + 6Cl (aq) + Cr 2 (CO 3 ) 3 (s) 3CO 3 2 (aq) + 2Cr 3+ (aq) Cr 2 (CO 3 ) 3 (s) Net ionic equation (b) : Ba(OH) 2 and HCl(aq) Ions: Ba 2+, OH, H +, Cl : BaCl 2 and H 2 O Solubility (Table 5.4, Table 3.10): Ba(OH) 2 (aq), HCl(aq), BaCl 2 (aq), H 2 O(l) Ba(OH) 2 (aq) + 2HCl(aq) BaCl 2 (aq) + 2H 2 O(l) Molecular equation Note that this is an acid-base reaction. The driving force for this reaction is the formation of water, which is a molecular compound. Next we write the ionic equation by separating all the aqueous ionic compounds and strong acids into their respective ions. Ba 2+ (aq) + 2OH (aq) + 2H + (aq) + 2Cl (aq) Ba 2+ (aq) + 2Cl (aq) + 2H 2 O(l) Finally, we write the net ionic equation by identifying and eliminating the spectator ions. Ba 2+ (aq) + 2OH (aq) + 2H + (aq) + 2Cl (aq) Ba 2+ (aq) + 2Cl (aq) + 2H 2 O(l) 2OH (aq) + 2H + (aq) 2H 2 O(l) We can reduce the coefficients by dividing each by 2. OH (aq) + H + (aq) H 2 O(l) Net ionic equation (c) : FeCl 3 and NaOH Ions: Fe 3+, Cl, Na +, OH : Fe(OH) 3 and NaCl Solubility (Table 5.4): FeCl 3 (aq), NaOH(aq), Fe(OH) 3 (s), NaCl(aq) FeCl 3 (aq) + 3NaOH(aq) Fe(OH) 3 (s) + 3NaCl(aq) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds into their respective ions. Fe 3+ (aq) + 3Cl (aq) + 3Na + (aq) + 3OH (aq) Fe(OH) 3 (s) + 3Na + (aq) + 3Cl (aq) Finally, we write the net ionic equation by identifying and eliminating the spectator ions. Fe 3+ (aq) + 3Cl (aq) + 3Na + (aq) + 3OH (aq) Fe(OH) 3 (s) + 3Na + (aq) + 3Cl (aq) Fe 3+ (aq) + 3OH (aq) Fe(OH) 3 (s) Net ionic equation (d) : Ca(HCO 3 ) 2 and HNO 3 Ions: Ca 2+, HCO 3, H +, NO 3 : Ca(NO 3 ) 2 and H 2 CO 3 which decomposes to CO 2 (g) and H 2 O(l)) Solubility (Table 5.4): Ca(HCO 3 ) 2 (aq), HNO 3 (aq), Ca(NO 3 ) 2 (aq); Acids are generally soluble. Other solubilities are obtained from Table 5.3. A reaction takes place because of the formation of liquid water and carbon dioxide gas Ca(HCO 3 ) 2 (aq) + 2HNO 3 (aq) Ca(NO 3 ) 2 (aq) + 2CO 2 (g) + 2H 2 O(l) Molecular equation Next we write the ionic equation by separating all the aqueous ionic compounds into their respective ions. Ca 2+ (aq) + 2 HCO 3 (aq) + 2H + (aq) + 2NO 3 (aq) Ca 2+ (aq) + 2NO 3 (aq)+ 2CO 2 (g) + 2H 2 O(l) Finally, we write the net ionic equation by identifying and eliminating the spectator ions. Ca 2+ (aq) + 2 HCO 3 (aq) + 2H + (aq) + 2NO 3 (aq) Ca 2+ (aq) + 2NO 3 (aq) + 2CO 2 (g) + 2H 2 O(l) 2 HCO 3 (aq) + 2H + (aq) 2CO 2 (g) + 2H 2 O(l) This coefficients can be reduced by dividing by 2. 5 31

HCO 3 (aq) + H + (aq) CO 2 (g) + H 2 O(l) Net ionic equation 5.113 Because we are starting with 2 reactants and forming 1 product, this is a combination reaction (Table 5.1). 5.114 Malachite s formula suggests that we consider the decomposition of each of its components. The decomposition of CuCO 3 produces the metal oxide and carbon dioxide gas (Table 5.2). The balanced equation for this reaction is: CuCO 3 (s) CuO(s) + CO 2 (g) The decomposition of Cu(OH) 2 produces copper oxide and water vapor. Cu(OH) 2 CuO(s) + H 2 O(g) Combining the two reactions we have: CuCO 3 Cu(OH) 2 (s) 2CuO(s) + CO 2 (g) + H 2 O(g) 5.115 For each reaction we: 1) determine the ion formulas, 2) predict the products based on ionic charges, and 3) determine the states for the products. If one or both products is insoluble, is a gas, or is a molecule (for example, water), a reaction takes place. (a) : ZnSO 4 (aq) and Ba(NO 3 ) 2 (aq) Ions: Zn 2+, SO 2 4, Ba 2+, NO 3 : BaSO 4 and Zn(NO 3 ) 2 Solubility (Table 5.4): BaSO 4 (s) and Zn(NO 3 ) 2 (aq) ZnSO 4 (aq) + Ba(NO 3 ) 2 (aq) BaSO 4 (s) + Zn(NO 3 ) 2 (aq) (b) : Ca(NO 3 ) 2 (aq) and K 3 PO 4 (aq) Ions: Ca 2+, NO 3, K +, PO 4 3- : Ca 3 (PO 4 ) 2 and KNO 3 Solubility (Table 5.4): Ca 3 (PO 4 ) 2 (s) and KNO 3 (aq) 3Ca(NO 3 ) 2 (aq) + 2K 3 PO 4 (aq) Ca 3 (PO 4 ) 2 (s) + 6KNO 3 (aq) (c) : ZnSO 4 (aq) and BaCl 2 (aq) Ions: Zn 2+, SO 4 2, Ba 2+, Cl : ZnCl 2 and BaSO 4 Solubility (Table 5.4): ZnCl 2 (aq) and BaSO 4 (s) ZnSO 4 (aq) + BaCl 2 (aq) BaSO 4 (s) + ZnCl 2 (aq) (d) : KOH(aq) and MgCl 2 (aq) Ions: K +, OH, Mg 2+, Cl : KCl and Mg(OH) 2 Solubility (Table 5.4): KCl(aq) and Mg(OH) 2 (s) 2KOH(aq) + MgCl 2 (aq) 2KCl(aq) + Mg(OH) 2 (s) (e) : CuSO 4 (aq) and BaS(aq) Ions: Cu 2+, SO 4 2, Ba 2+, S 2 : CuS and BaSO 4 Solubility (Table 5.4): CuS(s) and BaSO 4 (s) CuSO 4 (aq) + BaS(aq) CuS(s) + BaSO 4 (s) 5.116 (a) NaOH is a base, so it will react with HCl to form a salt and water in an acid-base reaction. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) (b) Zn is a metal that is more active than H 2. Therefore, it reacts with HCl to form an ionic compound and hydrogen gas in a single displacement reaction. Water is not formed during the reaction. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) 5 32

(c) CaCO 3 reacts with HCl to form carbonic acid, which decomposes to form water and carbon dioxide gas. CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 CO 3 (aq) CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) H 2 CO 3 decomposes (d) Cu(OH) 2 is a base, so it will react with HCl to form a salt and water in an acid-base reaction. Cu(OH) 2 (s) + 2HCl(aq) CuCl 2 (aq) + H 2 O(l) 5.117 Potassium and sodium are in the same family (Group IA (1)), so we expect that they will react similarly with water. Water reacts with potassium to form potassium hydroxide and hydrogen gas. 2K(s) + 2H 2 O(l) 2KOH(aq) + H 2 (g) 5.118 Both are solids prior to adding to water. Let s describe the dissolving of NaCl in detail. The solid is placed in the water. As the solid dissolves, the Na + and Cl ions separate and become aqueous. For every formula unit of NaCl that dissolves one sodium ion and one chloride ion form. Overall, the solution remains neutral. As Cu(NO 3 ) 2 dissolves, the Cu 2+ and NO 3 ions separate and become aqueous. For every formula unit of Cu(NO 3 ) 2 that dissolves, one copper ion and two nitrate ions form. Although the numbers of ions that form are different, the overall charge in the solution is still balanced. Because compounds containing nitrate and/or chloride ions are generally soluble, we don t expect to see any precipitation from the solution. 5.119 A precipitate of silver chloride will form if a compound which forms chloride ions is added to the solution. (a) Molecular chlorine does not form chloride ions when placed in water, so we don t expect to see a precipitate. (b) When sodium chloride is added to water, it dissolves, forming sodium and chloride ions. A precipitate of AgCl(s) should form. (c) Lead chloride is not soluble in water, so we cannot expect there to be enough Cl ions added to the solution to cause AgCl to precipitate. (d) Carbon tetrachloride is a molecular compound; it does not form chloride ions when added to water. The only molecular compounds that produce ions are acids and bases. 5.120 We could carefully add sodium sulfate, Na 2 SO 4, to the solution. As we add this solution, we would expect the following precipitation reaction to take place: BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) Molecular equation Since the sulfate ions precipitate out of the solution, the end result is a solution composed of sodium and chloride ions. This is clearer if we develop the net ionic equation: Ba 2+ (aq) + 2Cl (aq) + 2Na + (aq) + SO 4 2 (aq) BaSO 4 (s) + 2Na + (aq) + 2Cl (aq) Next we identify and eliminate the spectator ions: Ba 2+ (aq) + 2Cl (aq) + 2Na + (aq) + SO 4 2 (aq) BaSO 4 (s) + 2Na + (aq) + 2Cl (aq) Finally, we write the net ionic equation: Ba 2+ (aq) + SO 4 2 (aq) BaSO 4 (s) Net ionic equation Ionic equation There are two important conclusions to make. First, the sulfate precipitates and does not remain in the solution. Second, the spectator ions, Na + and Cl, remain in the solution. As a result, no new ions remain in the solution following precipitation. 5.121 Certain combinations of strong acid and strong base produce the net ionic equation: H + (aq) + OH (aq) H 2 O(l) Some examples are solutions of HCl and NaOH, HNO 3 and KOH, HCl and KOH. This reaction occurs because strong acids and bases completely ionize in aqueous solution. For example: 5 33

HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Molecular equation H + (aq) + Cl (aq) + Na + (aq) + OH (aq) Na + (aq) + Cl (aq) + H 2 O(l) H + (aq) + OH (aq) H 2 O(l) Net ionic equation Ionic equation 5.122 Any metal that is higher than cobalt in the activity series (Figure 5.21) can be used to plate cobalt from solution. However, if we use metals that are too active, they will also react with water (which is not very efficient). Good choices would be iron, zinc, or aluminum. 5.123 No reaction takes place because copper is less active than iron (Figure 5.21). 5.124 We can look at Table 5.4 to find an anion that will make an insoluble compound with lead(ii) but be soluble with copper(ii). Most sulfates are soluble but lead(ii) sulfate is insoluble. If we add sodium sulfate, the lead(ii) ions will precipitate as lead(ii) sulfate leaving the copper(ii) ions in solution. Next we look for an insoluble compound of copper(ii). There are several including copper(ii) hydroxide. Adding sodium hydroxide to the solution will precipitate the copper(ii) ions as copper(ii) hydroxide. 5.125 To show that KI is a strong electrolyte, we must show that it is an ionic solid that separates into its ions in solution. To show that CO is a nonelectrolyte, we must show that the CO molecules do not form ions in solution. Some representative images are shown below. Solution of KI Solution of CO K +, I CO 5.126 See Table 5.2. (a) Decomposition reactions can be used to produce oxygen gas. 2HgO(s) 2Hg(l) + O 2 (g) 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) (b) Single displacement reactions with very active metals with water produce hydrogen gas. 2Li(s) + 2H 2 O(l) 2LiOH(aq) + H 2 (g) (c) Decomposition of metal chlorides can be used to produce chlorine gas. PtCl 4 (s) Pt(s) + 2Cl 2 (g) 5.127 We can select any metal higher in the activity series (Figure 5.21). (a) Al; (b) Zn; (c) Mg; (d) Sn 5.128 We can identify the solutions by testing them with various active metals. Lead has the lowest activity. When we place a drop of each solution on a piece of lead foil, only the copper sulfate solution will react. Lead nitrate does not react. Next, we can place a drop of the three remaining solutions on a piece of tin foil (Sn). Of the three remaining solutions, only the lead nitrate solution will react. Finally, we can place a 5 34

drop of each of the two remaining solutions on a piece of zinc. Only the iron sulfate solution will react. The solution that remains is aluminum sulfate. 5.129 We can apply the solubility rules (Table 5.4) to separate ions. (a) Adding a potassium chromate solution will cause iron(iii) chromate to precipitate, leaving Al 3+ in solution. (b) Adding a sodium sulfate solution will cause barium sulfate to precipitate, leaving Mg 2+ in solution. (c) Adding a silver perchlorate solution will cause silver chloride to precipitate, leaving NO 3 in solution. (d) Adding water will cause MgSO 4 to dissolve, separating it from solid barium sulfate. 5.130 Copper probably reacts with oxygen to form copper(ii) oxide as described by the following equation: 2Cu(s) + O 2 (g) 2CuO(s) In the presence of water, the oxide forms the hydroxide: CuO(s) + H 2 O(l) Cu(OH) 2 (s) The reaction of CuO with CO 2 produces CuCO 3 : CuO(s) + CO 2 (g) CuCO 3 (s) 5.131 We can look at Table 5.4, the solubility rules, to find an anion that will make an insoluble compound with Ag + but be soluble with Ba 2+. Most chlorides are soluble but AgCl is insoluble. If we add sodium chloride (NaCl), the Ag + ions will precipitate as AgCl leaving the Ba 2+ ions in solution. Next we look for an insoluble compound of Ba 2+. There are several, including BaSO 4. Adding sodium sulfate (Na 2 SO 4 ) to the solution will precipitate the Ba 2+ ions as BaSO 4. 5.132 The reaction of a base, in this case Mg(OH) 2, with an acid, HCl, is a double- displacement, neutralization reaction. The products are a salt (i.e. ionic compound) and water. Just as with precipitation reactions we: 1) determine the ion formulas, 2) predict the products based on ion charges, and 3) determine the states of the products. Acid-base reactions are usually driven by the formation of water (a molecular compound). We must use the solubility rules to determine whether or not the salt will dissolve in water. : Mg(OH) 2 (s) and HCl(aq) Ions: Mg 2+, OH, H +, Cl : MgCl 2 (aq) and H 2 O(l) equation: Mg(OH) 2 (s) + 2HCl(aq) MgCl 2 (aq) + 2H 2 O(l) 5.133 In a single-displacement reaction, one free element displaces another element from a compound. For example, a metal displaces an ion of another metal, or a nonmetal displaces an ion of another nonmetal. To determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the metal whose ion appears in the compound, a reaction occurs. Nickel is more active than copper, so we expect a reaction to occur. In the lab we would observe the formation of copper on the surface of the nickel metal. The Cu 2+ ions in solution give it a blue color while nickel ions give no color to the solution. As the Cu 2+ ions react to form solid copper, the blue color of the solution would become lighter. As the nickel atoms form nickel ions they are removed from the surface of the metal. Over time we might observe that the nickel metal will appear to disintegrate. 5.134 In a single-displacement reaction, one free element displaces another element from a compound. For example, a metal displaces an ion of another metal, or a nonmetal displaces an ion of another nonmetal. To determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the metal whose ion appears in the compound, a reaction occurs. Silver is less reactive than copper, so we expect no reaction to occur. 5.135 Hydrochloric acid and sodium hydroxide solutions are both clear and colorless. They react in a doubledisplacement, neutralization reaction in which aqueous sodium chloride and water are formed. Aqueous 5 35

sodium chloride is also clear and colorless so there wouldn t be an observable change unless we inserted a thermometer. We would notice a change in temperature due to the formation of water. The formation of more water molecules in an environment that already has a lot of water would not produce a noticeable change in volume. 5.136 Using the solubility rules (Table 5.4) as a guide, we could identify a soluble compound containing the barium ion and one containing the sulfate ion such that when combined they would react in a doubledisplacement, precipitation reaction in which insoluble barium sulfate is formed along with a soluble compound. Since all nitrates are soluble, an aqueous solution of barium nitrate could serve as a source of barium ion. Sodium sulfate could serve as a source of sulfate ion. Mixing the two solutions would produce solid barium sulfate and aqueous sodium nitrate: Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaNO 3 (aq) The solid barium sulfate can be separated from the solution by filtration. 5.137 We start by writing the skeletal equation: C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Unbalanced 3 C 8 H 2 O 1 C 2 H 3 O In the combustion reactions of hydrocarbons it is usually easiest to begin by balancing the carbon and leave the balancing of oxygen for the last step. This follows the general concepts of balancing the atoms that appear in the smallest number of substances first (you could have also started with hydrogen) and balancing the substances appearing as pure elements last. Start with a coefficient of 3 in front of CO 2. C 3 H 8 (g) + O 2 (g) 3CO 2 (g) + H 2 O(g) Unbalanced 3 C 8 H 2 O 3 C 2 H 5 O To balance the hydrogen, we use a coefficient of 4 for H 2 O C 3 H 8 (g) + O 2 (g) 3CO 2 (g) + 4H 2 O(g) Unbalanced 3 C 8 H 2 O 3 C 8 H 10 O To balance the oxygen, we use a coefficient of 5 for O 2 C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) 3 C 8 H 10 O 3 C 8 H 10 O 5.138 In a single-displacement reaction, one free element displaces another element from a compound. For example, a metal displaces an ion of another metal, or a nonmetal displaces an ion of another nonmetal. In this case a metal displaces hydrogen from water producing an ionic compound and elemental hydrogen. To write a balanced equation we: 1) determine the ion formulas if ionic compounds are involved, 2) predict the products based on ion charges if applicable, and 3) determine the states of the products. : Mg(s) and H 2 O(l) Ions produced: Mg 2+, OH : Mg(OH) 2 (s) and H 2 (g) equation: Mg(s) + 2H 2 O(l) Mg(OH) 2 (s) + H 2 (l) 5 36

5.139 During this reaction atoms of elemental copper with a charge of zero are converted to Cu 2+ ions. Each copper atom that reacts loses two electrons. As a copper atom loses electrons, Ag + ions are converted to neutral silver atoms. Each silver ion that reacts gains one electron. 5.140 Barium sulfate is insoluble in water according to the solubility rules (Table 5.4). As an insoluble compound, toxic barium ions are not freely moving through our bodies to be absorbed into cells. Instead, the compound passes through the body with very little residual barium left behind. 5.141 (a) C 4 H 10 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Start with an atom count: : 4 C, 10 H, 2 O : 1 C, 2 H, 3 O First balance the carbon atoms by placing a coefficient in front of carbon dioxide: C 4 H 10 (g) + O 2 (g) 4CO 2 (g) + H 2 O(g) New atom count: : 4 C, 10 H, 2 O : 4 C, 2 H, 9 O Now balance the hydrogen atoms by changing the coefficient of water: C 4 H 10 (g) + O 2 (g) 4CO 2 (g) + 5H 2 O(g) New atom count: : 4 C, 10 H, 2 O : 4 C, 10 H, 13 O Now balance the oxygen atoms by changing the coefficient of oxygen: C 4 H 10 (g) + 13/2O 2 (g) 4CO 2 (g) + 5H 2 O(g) New atom count: : 4 C, 10 H, 13 O : 4 C, 10 H, 13 O Finally, we multiply through by 2 to remove the fractions: 2C 4 H 10 (g) + 13O 2 (g) 8CO 2 (g) + 10H 2 O(g) New atom count: : 8 C, 20 H, 26 O : 8 C, 20 H, 26 O The equation is now balanced. (b) C 2 H 5 OH(l) + O 2 (g) CO 2 (g) + H 2 O(g) Start with an atom count: : 2 C, 6 H, 3 O : 1 C, 2 H, 3 O First balance the carbon atoms by placing a coefficient in front of carbon dioxide: C 2 H 5 OH(l) + O 2 (g) 2CO 2 (g) + H 2 O(g) New atom count: : 2 C, 6 H, 3 O : 2 C, 2 H, 5 O Now balance the hydrogen atoms by changing the coefficient of water: C 2 H 5 OH(l) + O 2 (g) 2CO 2 (g) + 3H 2 O(g) New atom count: : 2 C, 6 H, 3 O : 2 C, 6 H, 7 O Now balance the oxygen atoms by changing the coefficient of oxygen: C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(g) New atom count: : 2 C, 6 H, 7 O : 2 C, 6 H, 7 O The equation is now balanced. (c) HCO 2 H(l) + O 2 (g) CO 2 (g) + H 2 O(g) Start with an atom count: : 1 C, 2 H, 4 O : 1 C, 2 H, 3 O The carbon and hydrogen atoms are balanced, so we balance the oxygen atoms by placing a coefficient in front of water: HCO 2 H(l) + O 2 (g) CO 2 (g) + 2H 2 O(g) New atom count: : 1 C, 2 H, 4 O : 1 C, 4 H, 4 O The hydrogen atoms are no longer balanced, so change the coefficient of formic acid: 2HCO 2 H(l) + O 2 (g) CO 2 (g) + 2H 2 O(g) New atom count: : 2 C, 4 H, 6 O : 1 C, 4 H, 4 O Now balance the carbon and oxygen atoms by changing the coefficient of carbon dioxide: 2HCO 2 H(l) + O 2 (g) 2CO 2 (g) + 2H 2 O(g) New atom count: : 2 C, 4 H, 6 O : 2 C, 4 H, 6 O The equation is now balanced. (d) C 5 H 12 (l) + O 2 (g) CO 2 (g) + H 2 O(g) Start with an atom count: : 5 C, 12 H, 2 O : 1 C, 2 H, 3 O First balance the carbon atoms by placing a coefficient in front of carbon dioxide: C 5 H 12 (l) + O 2 (g) 5CO 2 (g) + H 2 O(g) 5 37

New atom count: : 5 C, 12 H, 2 O : 5 C, 2 H, 11 O Now balance the hydrogen atoms by changing the coefficient of water: C 5 H 12 (l) + O 2 (g) 5CO 2 (g) + 6H 2 O(g) New atom count: : 5 C, 12 H, 2 O : 5 C, 12 H, 16 O Now balance the oxygen atoms by changing the coefficient of oxygen: C 5 H 12 (l) + 8O 2 (g) 5CO 2 (g) + 6H 2 O(g) New atom count: : 5 C, 12 H, 16 O : 5 C, 12 H, 16 O The equation is now balanced. (e) CH 3 OCH 3 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Start with an atom count: : 2 C, 6 H, 3 O : 1 C, 2 H, 3 O First balance the carbon atoms by placing a coefficient in front of carbon dioxide: CH 3 OCH 3 (g) + O 2 (g) 2CO 2 (g) + H 2 O(g) New atom count: : 2 C, 6 H, 3 O : 2 C, 2 H, 5 O Now balance the hydrogen atoms by changing the coefficient of water: CH 3 OCH 3 (g) + O 2 (g) 2CO 2 (g) + 3H 2 O(g) New atom count: : 2 C, 6 H, 3 O : 2 C, 6 H, 7 O Now balance the oxygen atoms by changing the coefficient of oxygen: CH 3 OCH 3 (g) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(g) New atom count: : 2 C, 6 H, 7 O : 2 C, 6 H, 7 O The equation is now balanced. CONCEPT REVIEW 5.142 Answer: D; both occur on the right side of the arrow so they are products A. both are reactants B. sulfuric acid is a reactant; iron(ii) sulfate is a product C. iron(ii) oxide is a reactant; water is a product E. iron(ii) oxide is a reactant; iron(ii) sulfate is a product 5.143 Answer: A; Conservation of matter requires that the total number of atoms does not change. B. 2H 2 (g) + O 2 (g) 2H 2 O(g) produces heat (Figure 5.5), so the temperature changes. C. Fe(s) + CuCl 2 (aq) FeCl 2 (aq) + Cu(s) changes the color of the solution from blue to colorless (Figure 5.2). D. 2H 2 (g) + O 2 (g) 2H 2 O(g) has three molecules as reactants and two molecules as products. E. 2HgO(s) 2Hg(l) + O 2 (g) has a solid changing to a liquid plus a gas. 5.144 Answer: C; the color change results from dilution of the green substance to give a lighter green. A. Based on the solubility rules, the white solid is AgCl: AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) B. The gray coating is silver metal and the blue solution contains copper(ii) nitrate: 2AgNO 3 (aq) + Cu(s) 2Ag(s) + Cu(NO 3 ) 2 (aq) D. The colorless gas is hydrogen (see the activity series in Figure 5.21): Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) E. This is a combination reaction, forming water: 2H 2 (g) + O 2 (g) 2H 2 O(g) 5.145 Answer: A; the image shows an equal number of the two reactants. B. There should be a 1:2 ratio of the two reactants. Possibilities are 1 N 2, 2 Cl 2 or 2 N 2, 4 Cl 2 or 3 N 2, 6 Cl 2, etc. 5 38

C. There should be a 1:2 ratio of the two reactants. Possibilities are 1 N 2, 2 O 2 or 2 N 2, 4 O 2 or 3 N 2, 6 O 2, etc. D. There should be a 1:3 ratio of the two reactants. Possibilities are 1 N 2, 3 H 2 or 2 N 2, 6 H 2 or 3 N 2, 9 H 2, etc. E. There should be a 1:3 ratio of the two reactants. Possibilities are 1 N 2, 3 O 2 or 2 N 2, 6 O 2 or 3 N 2, 9 O 2, etc. 5.146 Answer: B; this is the correct formula for sodium chloride and the atoms are balanced. A. The formula for sodium chloride is incorrect since the sodium ion has a charge of +1 and the chloride ion has a charge of 1. The equation is not balanced. C. The formula for sodium chloride is incorrect since the sodium ion has a charge of +1 and the chloride ion has a charge of 1. The equation is not balanced. D. The formula for sodium chloride is incorrect since the sodium ion has a charge of +1 and the chloride ion has a charge of 1. The equation is not balanced. E. The equation is balanced, but the formula for sodium chloride is not the simplest formula. 5.147 Answer: E since a compound is converted to another compound and an element. The reverse of this reaction would be a combination reaction. 5.148 Answer D; a metal reacting with a metal compound can potentially undergo a single-displacement reaction; the activity series in Figure 5.21 indicates that magnesium is a more active metal than nickel, so the reaction should occur. A. The formulas of these two compounds are incorrect; the common charge of these metal ions is +2, not +1. B. It would be impossible to balance the reaction with these products, since there would be more ionic charge on one side of the equation than on the other. C. If the nickel(ii) ion is going to change to the metal, the magnesium must change to its cation. E. The arguments for the correct answer eliminate this possibility. 5.149 Answer: A, B, and D; all of these choices represent the reaction of a substance with oxygen gas. C. Either change S to O 2 or change H 2 to O 2. E. Either change H 2 to O 2 or change Cl 2 to O 2. 5.150 Answer: D since CaCO 3 is an insoluble compound according to the solubility rules. A. Lithium carbonate is soluble, while calcium carbonate is insoluble, so this does not represent the chemical change. B. Lithium chloride is soluble, while calcium carbonate is insoluble, so this does not represent the chemical change. In addition, the equation does not use the smallest possible coefficients. C. Lithium chloride is soluble, while calcium carbonate is insoluble, so this does not represent the chemical change. E. The equation is balanced, but it is a molecular equation, not a net ionic equation. 5.151 Answer: A since the net ionic equation is Br (aq) + Ag + (aq) AgBr(s). B. Br is not a spectator ion since it is involved in the formation of the precipitate. C. Ag + is not a spectator ion since it is involved in the formation of the precipitate. D. Br is not a spectator ion since it is involved in the formation of the precipitate. E. NO 3 is a spectator ion, but this answer ignores that Co 2+ is also a spectator ion. 5 39