1. 12.3 Limiting Reagent and Percent Yield > 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations 12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield

2. 12.3 Limiting Reagent and Percent Yield > 2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents How is the amount of product in a reaction affected by an insufficient quantity of any of the reactants?

3. 12.3 Limiting Reagent and Percent Yield > 3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

4. 12.3 Limiting Reagent and Percent Yield > 4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Glossary Terms limiting reagent: any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction excess reagent: a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction

5. 12.3 Limiting Reagent and Percent Yield > 5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu(s) + S(s)  Cu 2 S(s) What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? Sample Problem 12.8 Determining the Limiting Reagent in a Reaction

6. 12.3 Limiting Reagent and Percent Yield > 6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant. KNOWNS mass of copper = 80.0 g Cu mass of sulfur = 25.0 g S UNKNOWN limiting reagent = ? Sample Problem 12.8

7. 12.3 Limiting Reagent and Percent Yield > 7 Start with one of the reactants and convert from mass of reactant to mass of product. Calculate Solve for the unknown. Sample Problem 12.8 63.5 g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S80.0 g Cu 1 mol Cu 2 S 159 g Cu 2 S = 100. g Cu 2 S Then, convert the other reactant to mass of the same product. 1 mol S 1 mol Cu 2 S25.0 g S 1 mol Cu 2 S 159 g Cu 2 S = 124 g Cu 2 S 1 mol S 32.1 g S Limiting Reagent Excess Reagent

8. 12.3 Limiting Reagent and Percent Yield > 8 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What is the maximum number of grams of Cu 2 S that can be formed when 80.0 g Cu reacts with 25.0 g S? 2Cu(s) + S(s)  Cu 2 S(s) Sample Problem 12.9 Using Limiting Reagent to Find the Quantity of a Product Cu is the Limiting Reagent so 100. g Cu 2 S is produced

9. 12.3 Limiting Reagent and Percent Yield > 9 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield What does the percent yield of a reaction measure?

10. 12.3 Limiting Reagent and Percent Yield > 10 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield What does the percent yield of a reaction measure? A batting average is actually a percent yield.

11. 12.3 Limiting Reagent and Percent Yield > 11 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield.

12. 12.3 Limiting Reagent and Percent Yield > 12 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield. The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.

13. 12.3 Limiting Reagent and Percent Yield > 13 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%. percent yield = actual yield theoretical yield  100%

14. 12.3 Limiting Reagent and Percent Yield > 14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. Percent Yield

15. 12.3 Limiting Reagent and Percent Yield > 15 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. Percent Yield The mass of the reactant is measured. The mass of one of the products, the actual yield, is measured. The percent yield is calculated. The reactant is heated.

16. 12.3 Limiting Reagent and Percent Yield > 16 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield Many factors cause percent yields to be less than 100%. Reactions do not always go to completion; when a reaction is incomplete, less than the calculated amount of product is formed. Impure reactants and competing side reactions may cause unwanted products to form. Actual yield can be lower than the theoretical yield due to a loss of product during filtration or in transferring between containers. If reactants or products have not been carefully measured, a percent yield of 100% is unlikely.

17. 12.3 Limiting Reagent and Percent Yield > 17 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is Sample Problem 12.10 Calculating the Theoretical Yield of a Reaction CaCO 3 (s)  CaO(s) + CO 2 (g)  What is the theoretical yield of CaO if 24.8 g CaCO 3 is heated?

18. 12.3 Limiting Reagent and Percent Yield > 18 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 Calculate the theoretical yield using the mass of the reactant. KNOWNS mass of CaCO 3 = 24.8 g CaCO 3 1 mol CaCO 3 = 100.1 g CaCO 3 (molar mass) 1 mol CaO = 56.1 g CaO (molar mass) 1 mol CaO/1 mol CaCO 3 (mole ratio from balanced equation) UNKNOWN theoretical yield = ? g CaO Sample Problem 12.10

19. 12.3 Limiting Reagent and Percent Yield > 19 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Start with the mass of the reactant and convert to moles of the reactant. Calculate Solve for the unknown. 2 Sample Problem 12.10 24.8 g CaCO 3  100.1 g CaCO 3 1 mol CaCO 3

20. 12.3 Limiting Reagent and Percent Yield > 20 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Next, convert to moles of the product using the mole ratio. Calculate Solve for the unknown. 2 Sample Problem 12.10 24.8 g CaCO 3   100.1 g CaCO 3 1 mol CaCO 3 1 mol CaO

21. 12.3 Limiting Reagent and Percent Yield > 21 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Finish by converting from moles to mass of the product. Calculate Solve for the unknown. 2 Sample Problem 12.10 24.8 g CaCO 3    100.1 g CaCO 3 1 mol CaCO 3 1 mol CaO 56.1 g CaO = 13.9 g CaO If there is an excess of a reactant, then there is more than enough of that reactant and it will not limit the yield of the reaction.

22. 12.3 Limiting Reagent and Percent Yield > 22 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO 3 is heated? Sample Problem 12.11 Calculating the Percent Yield of a Reaction CaCO 3 (s)  CaO(s) + CO 2 (g)  Calculate the theoretical yield first. Then you can calculate the percent yield.

23. 12.3 Limiting Reagent and Percent Yield > 23 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 Use the equation for percent yield. The theoretical yield for this problem was calculated in Sample Problem 12.10. UNKNOWN percent yield = ? % Sample Problem 12.11 KNOWNS actual yield = 13.1 g CaO theoretical yield = 13.9 g CaO (from sample problem 12.10) percent yield = actual yield theoretical yield  100%

24. 12.3 Limiting Reagent and Percent Yield > 24 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Substitute the values for actual yield and theoretical yield into the equation for percent yield. Calculate Solve for the unknown. 2 Sample Problem 12.11 percent yield =  100% = 94.2% 13.1 g CaO 13.9 g CaO

25. 12.3 Limiting Reagent and Percent Yield > 25 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. In this example, the actual yield is slightly less than the theoretical yield. Therefore, the percent yield should be slightly less than 100%. Evaluate Does the result make sense? 3 Sample Problem 12.11

26. 12.3 Limiting Reagent and Percent Yield > 26 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Carbon tetrachloride, CCl 4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is CS 2 + 3Cl 2  CCl 4 + S 2 Cl 2 What is the percent yield of CCl 4 if 617 kg is produced from the reaction of 312 kg of CS 2 ?

27. 12.3 Limiting Reagent and Percent Yield > 27 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. CS 2 + 3Cl 2  CCl 4 + S 2 Cl 2 What is the percent yield of CCl 4 if 617 kg is produced from the reaction of 312 kg of CS 2 ? 3.12  10 5 g CS 2    76.142 g CS 2 1 mol CS 2 1 mol CCl 4 1 mol CS 2 153.81 g CCl 4 1 mol CCl 4 = 6.30  10 5 g CCl 4 = 630 kg CCl 4 Percent yield =  100% = 97.9% 617 kg CCl 4 630 kg CCl 4

28. 12.3 Limiting Reagent and Percent Yield > 28 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. END OF 12.3