1. Percentage Yield

2. Assignment: Use Chapter 8 • Define the following terms: yield, theoretical yield, actual yield, percentage yield. • Based on your reading, give 4 reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. • Read the sample problem on the next slide and try the practice problem on slide number 5 • When 5.00 g of KClO3 is heated it decomposes according to the equation: 2KClO3 2KCl + 3O2 a) Calculate the theoretical yield of oxygen. b) Give the % yield if 1.78 g of O2 is produced. c) How much O2 would be produced if the percentage yield was 78.5%?

3. actual yield theoretical yield x 100% Answers 1) Yield: the amount of product Theoretical yield: the amount of product we expect, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is given in the question) Percent yield: • 2) • Not all product is recovered (e.g. spattering) • Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk) • A side reaction occurs (e.g. MgO vs. Mg3N2) • The reaction does not go to completion

4. 1 mol H2 actual 2 mol H2O 138 g H2O x = x 2.02 g H2 theoretical 143 g H2O 2 mol H2 18.02 g H2O x 1 mol H2O = = 143 g 96.7% Sample problem Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2 2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: # g H2O= 16 g H2 Step 3: Calculate % yield % yield = x 100% x 100%

5. actual 2 mol NH3 40.5 g NH3 x = theoretical 3 mol H2 227 g NH3 17.04 g NH3 x 1 mol NH3 = = 17.8% 227 g Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2 2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: # g NH3= 20.0 mol H2 Step 3: Calculate % yield % yield = x 100% x 100%

6. actual actual 1 mol KClO3 3 mol O2 78.5% x 1.958 g O2 x g O2 1.78 g O2 x = x = = 122.55 g KClO3 theoretical theoretical 1.958 g O2 1.958 g O2 100% 2 mol KClO3 32 g O2 x 1 mol O2 = = = = 78.5% 90.9% 1.537 g O2 1.958 g Answers # g O2= (also works if you use mol O2) 5.00 g KClO3 4) 2KClO3 2KCl + 3O2 a) b) c) % yield = x 100% x 100% % yield = x 100% x 100% x g O2

7. 1 mol O2 actual 1 mol H2 2 mol H2O 2 mol H2O 58 g H2O x x x x = 32 g O2 2.02 g H2 theoretical 1 mol O2 2 mol H2 62.4 g H2O 18.02 g H2O 18.02 g H2O x x 1 mol H2O 1 mol H2O = = 92.9% 68 g = 62.4 g Challenging question 2H2 + O2 2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first # g H2O= 60 g O2 # g H2O= 7.0 g H2 % yield = x 100% x 100%

8. More Percent Yield Questions Note: try “shortcut” for limiting reagent problems • The electrolysis of water forms H2 and O2. 2H2O  2H2 + O2 What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? • 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3 2KCI + 3O2 • What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

9. More Percent Yield Questions • Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? • 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2

10. 1 mol H2O actual 1 mol O2 12.3 g O2 x = x 18.02 g H2O theoretical 12.43 g O2 2 mol H2O 32 g O2 x 1 mol O2 = = 12.43 g 98.9% Q1 • The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? • Actual yield is given: 12.3 g O2 • Next, calculate theoretical yield # g O2= 14.0 g H2O Finally, calculate % yield % yield = x 100% x 100%

11. 1 mol KClO3 actual 3 mol O2 107 g O2 x = x 122.55 g KClO3 theoretical 117.5 g O2 2 mol KClO3 32 g O2 x 1 mol O2 = = 117.5 g 91.1% Q2 • 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3 2KCI + 3O2 • Actual yield is given: 107 g O2 • Next, calculate theoretical yield # g O2= 300 g KClO3 Finally, calculate % yield % yield = x 100% x 100%

12. actual 1 mol FeS 220 g O2 x = theoretical 1 mol Fe 263.7 g O2 87.91 g FeS x 1 mol FeS = = 83.4% 263.7 g Q3 • What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS • Actual yield is given: 220 g FeS • Next, calculate theoretical yield # g FeS= 3.00 mol Fe Finally, calculate % yield % yield = x 100% x 100%

13. actual 143 g Fe2O3 = theoretical 181.48 g Fe2O3 = = = 181.48 g Fe2O3 78.8% 199.7 g Fe2O3 1 mol O2 1 mol FeS2 2 mol Fe2O3 2 mol Fe2O3 159.7 g Fe2O3 159.7 g Fe2O3 x x x x x x 32 g O2 119.97 g FeS2 4 mol FeS2 11 mol O2 1 mol Fe2O3 1 mol Fe2O3 • 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 300 g FeS2 200 g O2 % yield = x 100% x 100%

14. actual 42% x 57.08 g Cl2 x g Cl2 = = theoretical 100% 57.08 g Cl2 = = = = 42% 57.08 g Cl2 24 g Cl2 62.13 g Cl2 1 mol MnO2 1 mol Cl2 1 mol Cl2 71 g Cl2 70.9 g Cl2 x x x x x 86.94 g MnO2 4 mol HCl 1 mol MnO2 1 mol Cl2 1 mol Cl2 # g Cl2= • 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 70 g MnO2 3.5 mol HCl % yield = x 100% x 100% x g Cl2 For more lessons, visit www.chalkbored.com