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. How many grams of Fe 2 O 3 are produced when 12.0 grams of iron rusts? 4Fe + 3O 2  2Fe 2 O 3 Do Now

3. 12.3 Limiting Reagent and Percent Yield > 3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. How many grams of Fe 2 O 3 are produced when 12.0 grams of iron rusts? 4Fe + 3O 2  2Fe 2 O 3 Do Now 159.8 g Fe 2 O 3 1 mol Fe 2 O 3 12.0 g Fe  55.9 g Fe 1 mol Fe  2 mol Fe 2 O 3 4 mol Fe  = 17.2 g Fe 2 O 3

4. 12.3 Limiting Reagent and Percent Yield > 4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 2 GC + 1 M + 4 Cp  1 Sm If I have:100 Graham Crackers 75 Marshmallows 250 Chocolate Pieces How many S’mores could we make? What limits how many we can make?

5. 12.3 Limiting Reagent and Percent Yield > 5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Glossary Terms theoretical yield: maximum amount of product that could be formed from given amounts of reactants 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

6. 12.3 Limiting Reagent and Percent Yield > 6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 2Cu + S  Cu 2 S If we have:100 moles Cu 75 moles S How many moles of Cu 2 S could we make? What limits how much we can make?

7. 12.3 Limiting Reagent and Percent Yield > 7 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

8. 12.3 Limiting Reagent and Percent Yield > 8 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Determining Limiting Reagent 1.Determine how much product you can make from each of the reactants. 2.The reactant that produces the smallest amount of product is the limiting reagent, and is completely consumed during the reaction. Sample Problem 12.8

9. 12.3 Limiting Reagent and Percent Yield > 9 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copper Determining Limiting Reagent Sample Problem 12.8 80.0 g Cu  63.5 g Cu 1 mol Cu  2 mol Cu 1 mol Cu 2 S Sulfur  1 mol S = 0.63 mol Cu 2 S = 0.78 mol Cu 2 S Copper is the limiting reagent.

10. 12.3 Limiting Reagent and Percent Yield > 10 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

11. 12.3 Limiting Reagent and Percent Yield > 11 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Convert from moles to mass of product. Sample Problem 12.9 159.1 g Cu 2 S 1 mol Cu 2 S 0.63 mol Cu 2 S  = 100 g Cu 2 S

12. 12.3 Limiting Reagent and Percent Yield > 12 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O 2 + 2H 2 O  2Fe(OH) 2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?

13. 12.3 Limiting Reagent and Percent Yield > 13 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Fe is the limiting reagent. 7.00 g Fe   = 0.13 mol Fe(OH) 2 1 mol Fe 55.85 g Fe2 mol Fe 2 mol Fe(OH) 2 Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O 2 + 2H 2 O  2Fe(OH) 2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? 9.00 g H 2 O   = 0.50 mol Fe(OH) 2 1 mol H 2 O 18 g H 2 O2 mol H 2 O 2 mol Fe(OH) 2

14. 12.3 Limiting Reagent and Percent Yield > 14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield theoretical yield : maximum amount of product that could be formed from given amounts of reactants. actual yield : amount of product that actually forms when the reaction is carried out in the laboratory.

15. 12.3 Limiting Reagent and Percent Yield > 15 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?

16. 12.3 Limiting Reagent and Percent Yield > 16 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 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

17. 12.3 Limiting Reagent and Percent Yield > 17 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.

18. 12.3 Limiting Reagent and Percent Yield > 18 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 12.11 percent yield =  100% = 94.2% 13.1 g CaO 13.9 g CaO

19. 12.3 Limiting Reagent and Percent Yield > 19 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 g is produced from the reaction of 312 g of CS 2 ?

20. 12.3 Limiting Reagent and Percent Yield > 20 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 g is produced from the reaction of 312 g of CS 2 ? 312 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 Theoretical yield = 630 g CCl 4 Percent yield =  100% = 97.9% 617 g CCl 4 630 g CCl 4

21. 12.3 Limiting Reagent and Percent Yield > 21 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO 2 + 3C  SiC + 2CO What is the percent yield of this reaction?

22. 12.3 Limiting Reagent and Percent Yield > 22 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO 2 + 3C  SiC + 2CO What is the percent yield of this reaction? 83.5%

23. 12.3 Limiting Reagent and Percent Yield > 23 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. END OF 12.3