CHAPTER 12 Stoichiometry 12.1 Using Everyday Equations Stoichiometry is the calculation of quantities in chemical equations. * The balanced equation gives the ratios for the reactants and products. 3 eggs + 2 cups Flour + 1 Cup sugar 1 cake How many eggs are needed to make 4 cakes? How many cups of flour are needed if you wanted to use 9 eggs? If you needed to get rid of 5 cups of sugar, how many cakes must you make? * The balanced equation gives the ratios for the reactants and products. * 1 mole of Mg reacts with 2 moles of HCl * 1 mole of MgCl 2 and 1 mole of H 2 are produced Mg + 2HCl MgCl 2 + H 2 How many moles of HCl are needed to react 2 moles of Mg? How many moles of H2 are produced if 6 moles of HCl reacts? How much Magnesium is needed to produce 7.6 moles of MgCl 2? Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. Interpreting Chemical Equations N 2 (g) + 3H 2 (g) 2NH 3 (g) Rep. PARTICLES MOLES MASS VOLUME 1
What is conserved in a Chemical Reaction? Atoms Atoms Mass Mass Moles Moles Molecules Molecules Volume Volume What is Conserved in Chemical Equations? Mass & atoms always! Volume is not because temp. or pressure can change. Moles, molecules, and formula units are not because atoms rearrange in a chemical reaction. Ratios in chemical equations What does excess mean? If you have excess reagent, that means you have plenty of it. Reagent = Reactant Ratios in chemical equations Can change but should be listed in lowest whole number form. Lowest whole number Ratios are used during mole to mole conversions. 3 eggs + 2 cups Flour + 1 Cup sugar 1 cake You have: > 12 eggs > 10 cups Flour > 8 cups sugar Which ingredient limits the number of cakes you can bake? Stoichiometric Pathways N 2 (g) + 3H 2 (g) 2NH 3 (g) How would you convert volume of nitrogen gas into mass of nitrogen trihydride? Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) How would you convert atoms of zinc into volume of hydrogen gas? Which ingredients do you have an excess of? How would you convert moles of Zinc into formula units of zinc chloride? 2
Comparing Atoms Comparing Molecules 1 mol H 2 O 1 mol H 2 O 2 6.02 x 10 23 molecules 6.02 x 10 23 molecules H: O: gmm = 18g H: O: gmm = 34g Comparing Ionic Compounds 1 mol CaCl 2 1 mol Al 2 (CO 3 ) 3 6.02 x 10 23 form units 6.02 x 10 23 form units Ca: Cl: gfm = 111 g Al: C: O: gfm = 234g 12.2 CHEMICAL CALCULATIONS Mole Mole Calculations Begin with a mole quantity End with a mole quantity One step operation 2KClO 3 2KCl + 3O 2 What mass of KCl is produced from 25.0 g of KClO 3? 25.0 grams KClO 3 Mole Mole Calculations N 2 (g) + 3H 2 (g) 2NH 3 (g) How many moles of ammonia can be produced from 0.60 mol of nitrogen gas? 3
More Mole Mole Calculations How many moles of Al is needed to make 3.7 mol of Al 2 O 3? More Mole Mole Calculations How many moles O 2 is needed to completely react with 1.48 x 10 1 mol of Al? More Mole Mole Calculations How many moles of Al 2 O 3 can be made from 7.8 x 10 1 mol O 2? Mass Mass Calculations 4 step process Mass Change mass to moles Mole mole ratio Change mole to mass More Mass Mass Calculations Grams of Al 2 O 3 from 80.0 g of O 2? Mass Mass Calculations How many grams of Al is needed to form 4.56 x 10 6 grams of Al 2 O 3 with an excess of oxygen? 80.0 grams O 2 4.56 X 10 6 grams Al 2O 3 4
Other Stoichiometric Calculations Mass Molar Road Map Mass Stoichiometry calculations may include any unit of measurement that is related to the mole. Volume MOL Volume Molar Road Map Rep. Particles Rep. Particles Other Stoichiometric Calculations 2H 2 O 2H 2 + O 2 How many molecules of oxygen are produced from 29.2 g of water? 2H 2 O 2H 2 + O 2 Assuming STP, how many liters of hydrogen are made from 4.32 x 10 24 molecules of water? 4.32 X 10 24 molecules H 2O 29.2 grams H 2O Assuming STP, how many liters of ammonia are produced from 7.03 x 10 25 molecules of hydrogen gas with an excess of nitrogen? Assuming STP, how many grams of aluminum are needed to produce 6.31 x 10 31 formula units of Aluminum oxide with an excess of oxygen? 6.31 X 10 31 Forms Al 2O 3 7.03 X 10 25 molecules H2 5
12.3 Limiting Reagent and Percent Yield Reactant that is used up first in a chemical reaction, or what you will run out of first. It determines the amount of product that can be formed. EXCESS REAGENT Reactant that you have more than enough of. This will remain unchanged in the equation. Example for Understanding You are making omelet's... Each omelet needs 3 eggs and 1 cup of cheese. You have 12 eggs and 10 cups of cheese. 3 eggs + 1 cups cheese 1 omelet How many omelets can you make? Which ingredient limits the number of omelets you can make? Which ingredient will you have an excess of? Determining Limiting Reagent Using the amounts of substances present, perform a mol mol conversion for each. This gives you the amount needed. Compare the amount needed with the amount present. Whichever substance has too little present is the limiting reagent. * when given both reactants Example #1: 2Na + Cl 2 2NaCl Given 6.70 mol of Na and 3.20 mol of Cl 2. Step 1. What is the limiting reagent? HAVE NEED 6.70 mol Na 1 mol Cl 2 2 mol Na 3.20 mol Cl 2 2 mol Na 2Na + Cl 2 2NaCl Given 6.70 mol of Na and 3.20 mol of Cl 2. Step 2: What is the excess reagent and how much excess is there? Sodium: 6.70 moles 6.40 moles= 0.3 moles in Excess 1 mol Cl 2 6
2Na + Cl 2 2NaCl Given 6.70 mol of Na and 3.20 mol of Cl 2. Step 3. How many moles of NaCl are made? 3.20 mol Cl 2 2 mol NaCl 1 mol Cl 2 LR determines Product made Example #2: 2Cu + S Cu 2 S Given 80.0 g of copper and 25.0 g of sulfur. Step 1. What is the limiting reagent? HAVE NEED 80.0 g Cu 1 mol Cu 1 mol S 32 g S 64 g Cu 2 mol Cu 1 mol S 25.0 g S 1 mol S 2 mol Cu 32 g S 1 mol S 64 g Cu 1 mol Cu 2Cu + S Cu 2 S Given 80.0 g of copper and 25.0 g of sulfur. Step 2: What is the excess reagent and how much excess is there? 2Cu + S Cu 2 S Given 80.0 g of copper and 25.0 g of sulfur. Step 3. How much Cu 2 S can be made? 80.0 g Cu 1 mol Cu 1 mol Cu 2 S 160 g Cu 2 S Sulfur: 25.0 grams 20.0 grams= 5.0 grams in Excess 64 g Cu 2 mol Cu 1 mol Cu 2 S This answer should be in scientific notation... Example #3: Given 341.0 L of N 2 and 236.0 L of H 2 @ STP Step 1. What is the limiting reagent? HAVE NEED Given 341.0 L of N 2 and 236.0 L of H 2 @ STP Step 2: What is the excess reagent and how much excess is there? 341.0 L N2 1 mol N2 3 mol H2 1 mol N2 1 mol H2 Nitrogen: 341.0 liters 78.67 liters= 262.3 liters in Excess 236.0 L H2 1 mol H2 1 mol N2 3 mol H2 1 mol N2 7
Given 341.0 L of N 2 and 236.0 L of H 2 @ STP Step 3. How many grams of NH 3 can be made? Example # 4: How many liters of NH 3 are produced if 85.0 grams of N 2 and 85.0 grams of H 2 react @ STP? 85.0 g N 2 1 mol N 2 3 mol H 2 2 g H 2 28 g N 2 1 mol N 2 1 mol H 2 236.0 L H 2 1 mol H 2 2 mol NH 3 3 mol H 2 17 g NH 3 1 mol NH 3 85.0 g H 2 1 mol H 2 1 mol N 2 28 g N 2 2 g H 2 3 mol H 2 1 mol N 2 This answer should be in scientific notation... 85.0 g N 2 1 mol N 2 2 mol NH 3 28 g N 2 1 mol N 2 1 mol NH 3 Percent Yield Theoretical yield: the maximum amount of product that can form from the amount of reactants given. > (Regular Stoichiometry) Actual yield: the amount of product that actually forms during a reaction. Cannot be greater than the theoretical yield. > (Often less) Percent yield: a percentage comparing the actual yield with the theoretical yield. (Same units) > % = (Actual Theoretical) x 100 Determining Percent Yield CaCO 3 CaO + CO 2 24.8 g of calcium carbonate is heated to produce an actual yield of 13.1 g of calcium oxide. 1. What is the theoretical yield? 24.8 g CaCO3 1 mol CaCO 3 1 mole CaO 100 g 1 mol CaCO 3 56 g 1 mol CaO Determining Percent Yield CaCO 3 CaO + CO 2 24.8 g of calcium carbonate is heated to produce an actual yield of 13.1 g of calcium oxide. 2. What is the percent yield? (Must be same units!!) Actual Yield Theoretical Yield x 100 = Determining Percent Yield 2CuCl + H 2 S Cu 2 S + 2HCl 44.0 mol of CuCl forms an actual yield of 6.0 mol of Cu 2 S 1. Theoretical yield of Cu 2 S in grams? 44.0 mol CuCl 1 mol Cu2S 160 g 2 mol CuCl 1 mol Cu 2S 8
Determining Percent Yield 2CuCl + H 2 S Cu 2 S + 2HCl 44.0 mol of CuCl forms an actual yield of 6.0 mol of Cu 2 S 2. Actual yield of Cu 2 S in grams? 6.0 mol Cu 2 S 160 g 1 mol Cu 2 S 2CuCl + H 2 S Actual Yield Theoretical Yield Cu 2 S + 2HCl 44.0 mol of CuCl forms an actual yield of 6.0 mol of Cu 2 S 3. What is the percent yield of Cu 2 S? (same units!) x 100 = 4NH 3 + 5O 2 4NO + 6H 2 O 9.0 g of NH 3 produces a percent yield of 63% water. 1. Theoretical yield of H 2 O in grams? 2. Actual yield of H 2 O in grams? Hint: Actual Theoretical =.63 4NH 3 + 5O 2 4NO + 6H 2 O 1. What is the theoretical yield of H 2 O in grams? 9.0 g NH 3 1 mol NH 3 6 mol H 2 O 17 g 4 mol NH3 18 g 1 mol H 2 O 2. What is the actual yield of H 2 O (in grams) if only 63% was actually produced? 2Na + Cl 2 2NaCl Given 4.50 mol of Na and 4.50 mol of Cl 2, how many grams of NaCl can be produced? HAVE 4.50 mol Na 4.50 mol Cl2 1 mol Cl2 2 mol Na 2 mol Na NEED CaCO 3 CaO + CO 2 31.7 g of calcium carbonate is heated to produce an actual yield of 13.1 g of calcium oxide. What is the percent yield of CaO? 31.7 g CaCO3 1 mol CaCO 3 1 mole CaO 100 g 56 g 1 mol CaCO 3 1 mol CaO 1 mol Cl2 9