Chapter 5 Chemical Reactions and Equations

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1 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 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 O

2 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 O 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 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 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

3 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 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) 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) 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 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 There is a rearrangement of the atoms into new substances. Since new substances are formed, a chemical reaction has taken place 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 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 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 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) 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

4 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 (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 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 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 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

5 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

6 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

7 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 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

8 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 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 (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

9 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

10 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

11 (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

12 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

13 (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 (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

14 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

15 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 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

16 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) 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) (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 (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 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

17 (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) 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 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

18 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 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) 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 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) 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

19 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 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

20 (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 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 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 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

21 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

22 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 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 From the problem description we can write: CaCl 2 (aq) + K 2 CO 3 (aq) 5 22

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