BALANCING CHEMICAL EQUATIONS

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1 BALANCING CHEMICAL EQUATIONS The Conservation of Matter states that matter can neither be created nor destroyed, it just changes form. If this is the case then we must account for all of the atoms in a chemical reaction. We cannot change the way compounds are put together but we can adjust the number of compounds that are made. For example: Magnesium reacts with oxygen to form magnesium oxide. Since oxygen is a diatomic molecule we can only obtain them in groups of two atoms. We know that the formula for magnesium oxide is MgO(s). The reaction as we know it so far is: Mg(s) + O2 (g) ---> MgO (s) What happens to the other oxygen? It must be accounted for! When the oxygen molecules breaks apart to react with the magnesium atom the lone oxygen is free to bond with another magnesium. In other words we have now used up two magnesium atoms and created two units of magnesium oxide. We show this by placing numbers in front of the molecule. 2 Mg(s) + O2 (g) ---> 2 MgO (s) This is a balanced equation. It can be read as follows: Two atoms of magnesium react with 1 molecule of oxygen to form two units of magnesium oxide. The goal is to ensure that the same number of atoms appear on each side of the equation! Look at the worksheet on page 1. Count the atoms to see if these equations are balanced. To balance an equation you may start with any atom but you may find it helpful to start with the one present in higher numbers. Example: Sulfur reacts with lithium to form lithium sulfide. S8 (s) + Li (s) ---> Li2S (s) Start with the sulfur atoms. 8 on the reactant side means we need 8 on the product side. We must place an 8 in front of the Li2S(s). This changes the number of lithium atoms so that we now have 16 on the product side. We must then have 16 on the reactant side. The equation becomes: S8 (s) + 16 Li (s) ---> 8 Li2S (s) Try these. 1. Sodium reacts with chlorine gas to form sodium chloride. Na(s) + Cl2 (g) ---> NaCl(s) Balanced equation: 2 Na (s) + Cl2 (g) ---> 2 NaCl (s) 2. Barium reacts with phosphorus to produce barium phosphide. Ba (s) + P4 (s) ---> Ba3P2 (s) Balanced equation: 6 Ba (s) + P4 (s) ---> 2 Ba3P2 (s)

2 Formation or Simple Composition Reactions Formation reactions are reactions where two elements combine to form a compound. element + element = compound Word equation: Skeleton equation: sodium + chlorine > sodium chloride Na(s) + Cl(g) > NaCl(s) Decomposition Reactions Decomposition reactions are the reverse of formation reactions. compound AB ---> element A + element B ex. C 6H12O6 (s) ---> _ C (s) + _ H2 (g) + _ O2 (g) Single Replacement Reactions So far we have dealt with formation and decomposition reactions. Elements react to form compounds or compounds react to form elements. Very few reactions are of this type. Reactions that involve an element and a compound can be much more interesting. These reactions are called single replacement reactions because one of the elements in the compound is replaced by another element. Example: copper reacts with silver nitrate solution to from silver and copper (II) nitrate solution. Cu(s) + AgNO3 (aq) ---> Ag (s) + Cu(NO3)2 (aq) You can balance these equations in the same manner that you did the formation reactions. ie. count the number of atoms of each type on each side of the equation. It is easier, however, to treat the nitrate ion as a single atom. This means you count Ag, Cu and NO3ʼs. The balanced equation is thus: Cu (s)+2 AgNO3 (aq) ---> 2 Ag (s) +Cu(NO3)2 (aq) Examples: 1. Calcium metal reacts with sodium carbonate solution to produce solid calcium carbonate and sodium metal. 2. Nitrogen gas reacts with ammonium phophide solution to produce a solution of ammonium nitride and solid phosphorus.

3 Double Replacement Reactions These reactions are similar to single replacement reactions. The only difference is that both the metal and nonmetal will be replaced in this reaction between two compounds. Example: Magnesium hydroxide tablets are used to neutralize stomach acid (HCl). This produces water and magnesium chloride solution. _Mg(OH)2 (s) +_HCl (aq)--->_hoh (l) + _MgCl2 (aq) Mg(OH) 2 has 2 OH ions where HOH only has 1 OH ion. We must have 2 OH on each side so we put a 2 in front of HOH to balance the OHʼs. This changes the H ions in HOH to 2. We need 2 Hʼs on the left side also. Put a two in front of the HCl. This gives us 2 Cl and MgCl2 has 2 so it balances. The Mgʼs also are balanced so we are done. Example: Sodium arsenide solution is mixed with calcium chloride solution to produce a solution of sodium chloride and a calcium arsenide solid. _Na3As (aq)+_cacl2 (aq)-->_nacl (aq) +_Ca3As2 (s) You may start with any element or complex ion you like, however it may be easier to start with the greatest number. In this example we can start with calcium or sodium as both have 3 on one side of the equation. Lets start with sodium since it is first. Na3As (aq) has 3 Na. We must therefore have 3 one the other side of the equation so we put a 3 in front of the NaCl (remember we cannot change the chemical formula only the number of formula units) This changes the number of chloride ions to 3 as well. That means we will have to have 3 chloride ions on the other side of the equation. Since chloride ions only are present in groups of 2 in CaCl2 we must find a way to get the chloride ions equal on each side. We require something in which 2 and 3 both can fit into. The smallest number that satisfies this is 6. Put a 3 in front of the CaCl2 to get 6 Cl ions. That means we must put a 6 in front of the NaCl. This changes the number of Naʼs to 6. To balance the left side we must put a 2 in front of the Na3As. We now have 6 Na on each side and 6 Cl. Check the remaining ions. Other Reaction Type When you have a polyatomic ion that breaks apart, or when you have more products than reactants in an reaction it is classified as other. With other reaction types do hydrogen atoms second last and then oxygen last. Why? Hydrogen and Oxygen atoms tend to balance themselves out when the other atoms are balanced. Ex. Al (s) + NH4ClO4 (s) --> Al2O3 (s) + AlCl3 (s) + NO (g) + HOH(l)

4 Precipitate- a solid formed from a solution in a reaction. If a compound has an aqueous phase on the reactant side and then becomes solid on the product side, it is a precipitate Ex. _NH4Br (aq) + Hg3N (aq) --> (NH4)3N (aq) + HgBr (s) Identify the precipitate in the above reaction? Hydrocarbon Combustion Reactions When hydrocarbons such as gasoline, methane, propane and sucrose are burned (combusted) they always produce energy plus carbon dioxide gas and water vapor. The act of burning is actually just a rapid reaction with oxygen. The equations are just the compound to be burned plus oxygen to produce carbon dioxide gas and water vapor. When balancing reactions of this type it is easiest to begin with the carbon atoms then move onto the hydrogen atoms and finish with the oxygen atoms. Example: Methane gas is burned. _ CH4 (g) + _O2 (g) --->_CO2 (g) + _H2O (g) Balance by starting with the carbon atoms. One on each side so no change is required. There are 4 hydrogen atoms on the left side and only 2 one the right side. Need 2 water molecules to get 4 hydrogen atoms. Next, count the oxygen atoms on the right side. We have 2 in CO2 and 2 in 2H2O. That makes 4 all together. To get 4 on the other side we require 2 O2. CH4 (g) + 2 O2 (g) ---> CO2 (g) + 2 H2O (g) Sometimes it is necessary to adjust the number of our starting compound. Example: Octane is combusted. _ C8H18 (l) +_O2 (g) --->_CO2 (g) + _H2O (g) Begin as before with the carbon and move on to the hydrogen. We have the following: _ C8H18 (l) +_O2 (g) --> 8 CO2 (g) + 9 H2O (g) Count the oxygen next. 16 in 8 CO2 and 9 in the 9 H2O. This makes 25. Since oxygen only comes in packages of 2 (O2) we cannot get 25! We need more oxygen on the left side. If we double everything we will have 50 oxygen on the right side. That is divisible by 2. We end up with: 2 C8H18 (l) + 25 O2 (g)--> 16 CO2 (g) +18 H2O (g)

5 Endothermic & Exothermic Reactions All reactions have energy in them, but most of the time we donʼt write it in the equation. You have seen some reactions where energy is written in the equation. Ex. Photosynthesis reaction. SolarEnergy+6CO 2 (g)+6h 2 O(l)-->6O 2 (g)+c 6 H 12 O 6 (s) Most reactions require energy to get the reactants to react. An input of energy. When there is more energy taken in (to start the reaction) than what is given off from the reaction it is called endothermic (ie. photosynthesis reaction). Energy is written on the reactant side in an endothermic reaction. When there is more energy given off in the reaction than what is put into the reaction it is called exothermic Energy is written on product side in an exothermic reactions. *Note- most of the time the energy in the reactions are in the form of heat, but not always. Evidence for Chemical Reactions There are four things that prove that a chemical reaction has occurred: 1. Temperature change 2. Color change 3. Gas given off 4. Precipitate is formed A chemical reaction can have more than one. The Mole The mole is a unit representing a certain number of things. Much the same as a dozen or a gross, but instead of being 12 or 144 things a mole is 6.02 x things. It is sometimes referred to as Avogadroʼs number and it is defined as the number of atoms in exactly 12 grams of carbon-12. Using The Mole All atomic masses are compared to carbon-12. The atomic mass number given on your periodic table is given to the nearest one-hundredth of a gram. Example: One mole of chlorine atoms has a mass of g/mol. Note: the symbol for mole is mol. This is not an abbreviation. Find: Oxygen, Calcium and Iron.

6 Molar Mass of Molecules Most often we are interested in a mole of molecules. That is 6.02 x 1023 molecules such as water, HOH. One molecule of water contains 2 atoms of hydrogen and one atom of oxygen. This means that one mole of water molecules contains 2 moles of hydrogen atoms and one mole of oxygen atoms. The mass of one mole of water therefore is the sum of 2 moles of hydrogen atoms and one mole of oxygen atoms. 2 H = 2 mol x 1.01g/mol = 2.02 g/mol 1 O= 1 mol x g/mol = g/mol 1 HOH = g/mol Example: Calculate the molar mass of hydrogen sulfate gas. Formula is H2SO4 (g) 2 H = 2 mol x 1.01g/mol = 2.02 g/mol 1 S = 1 mol x 32.06g/mol = g/mol 4 O = 4 mol x g/mol = g/mol 1 mole of H2SO4 (g) = g/mol Try: calcium chloride : iron (III) hydroxide : phosphorus pentachloride Converting Mass to Moles The number of moles (n) present in a given mass (m) is equal to the given mass divided by the molar mass (M). n= m M Example 1: How many moles of sodium chloride are present in 100 g? Step 1: Calculate the molar mass of sodium chloride. MNaCl = g/mol Step 2: Calculate the number of moles. n= m M = 100 g g/mol = 1.71 mol Remember your significant digits!!!!!

7 Example 2: How many moles are present in 25 g of aluminum nitrate? Answer = 0.12 mol Moles to Mass Calculations To calculate the mass present in a given number of moles we need to rearrange our formula as follows. m= nm Example: What mass is present in 3.50 mol of calcium carbonate? Step 1: Calculate the molar mass of calcium carbonate. MCaCO3 = g/mol Step 2: Calculate the mass. m = nm = 3.50 mol x g/mol = 350 g Example 2: Calculate the mass of mol of ammonium sulfide. Answer = 3.8 g

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