1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 9 Chemical Quantities in Reactions On a space shuttle, the LiOH in the canisters removes CO 2 from the air. 9.3 Limiting Reactants

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product calculated using the balanced equation. Actual yield The amount of product obtained when the reaction takes place. Percent yield The ratio of actual yield to theoretical yield. Percent yield = actual yield (g) x 100 theoretical yield (g)

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Suppose you prepare cookie dough to make 5 dozen cookies. The phone rings and you answer. While talking, a sheet of 12 cookies burns and you have to throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100% = 80.% yield 60 cookies Calculating Percent Yield

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Calculations for Percent Yield

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 Learning Check Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O 2 (g) 2CO(g) What is the percent yield if 40.0 g of CO is produced when 30.0 g of O 2 is used? 1) 25.0% 2) 75.0% 3) 76.2%

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Solution STEP 1 Write the given and needed quantities. Given 40.0 g of CO produced (actual) 30.0 g of O 2 used Need percent yield of CO STEP 2 Write a plan to calculate the theoretical yield and the percent yield. g of O 2 moles of moles of g of CO O 2 CO (theoretical) Percent yield of CO = g of CO (actual) x 100% g of CO (theoretical)

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Solution (continued) STEP 3 Write the molar masses and the mole-mole factors from the balanced equation. 1 mol of O 2 = 32.00 g of O 2 1 mol O 2 and 32.00 g O 2 32.00 g O 2 1 mol O 2 1 mol of O 2 = 2 mol of CO 1 mol O 2 and 2 mol CO 2 mol CO 1 mol O 2 1 mol of CO = 28.01 g of CO 1 mol CO and 28.01 g CO 28.01 g CO 1 mol CO

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 Solution (continued) STEP 4 Calculate the percent yield ratio by dividing the actual yield (given) by the theoretical yield x 100%. 30.0 g O 2 x 1 mol O 2 x 2 mol CO x 28.01 g CO 32.00 g O 2 1 mol O 2 1 mol CO = 52.5 g of CO (theoretical) Percent yield: 40.0 g CO (actual) x 100 = 76.2% yield (3) 52.5 g CO (theoretical)

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 Learning Check When N 2 and 5.00 g of H 2 are mixed, the reaction produces 16.0 g of NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 1) 31.3 % 2) 56.7 % 3) 80.0 %

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Solution STEP 1 Write the given and needed quantities. Given 16.0 g of NH 3 produced (actual) 5.00 g of H 2 Need percent yield of NH 3 STEP 2 Write a plan to calculate the theoretical yield and the percent yield. g of H 2 moles of H 2 moles of NH 3 g of NH 3 (theoretical) Percent yield = g of NH 3 (actual) x 100% g of NH 3 (theoretical)

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Solution (continued) STEP 3 Write the molar masses and the mole- mole factors from the balanced equation. 1 mol of H 2 = 2.016 g of H 2 1 mol H 2 and 2.016 g H 2 2.016 g H 2 1 mol H 2 3 mol of H 2 = 2 mol of NH 3 3 mol H 2 and 2 mol NH 3 2 mol NH 3 3 mol H 2 1 mol of NH 3 = 17.03 g of NH 3 1 mol NH 3 and 17.03 g NH 3 17.03 g NH 3 1 mol NH 3

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 Solution (continued) STEP 4 Calculate the percent yield ratio by dividing the actual yield (given) by the theoretical yield x 100%. 5.00 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17.03 g NH 3 2.016 g H 2 3 mol H 2 1 mol NH 3 = 28.2 g of NH 3 (theoretical) Percent yield = 16.0 g NH 3 x 100 = 56.7 % (2) 28.2 g NH 3