Do Now 1 Read the balanced equation below and write all possible mole ratios. 2Al 2 O 3 → 4Al(s) + 3O 2 (g)

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Presentation transcript:

Do Now 1 Read the balanced equation below and write all possible mole ratios. 2Al 2 O 3 → 4Al(s) + 3O 2 (g)

2 Chapter 9 Percentage Yield Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Essential Question 3 What is the difference between the theoretical yield and the actual yield?

4 Theoretical, Actual, and Percent Yield T heoretical yield The maximum amount of product calculated using the balanced equation. (mass of product that should be produced from the limiting reactant) Actual yield The amount of product obtained when the reaction takes place. (mass of product actually produced from the limiting reactant) Percent yield The ratio of actual yield to theoretical yield. percent yield = actual yield (g) x 100 theoretical yield (g)

5 To calculate the percent yield, the actual yield and theoretical yield are needed. You prepared cookie dough using a recipe to make 5 dozen cookies. After all of the preparation, you only end up with 48 cookies. What is the percent yield of cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies What are some possible sources of error? Calculating Percent Yield

6 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 CO are produced when 30.0 g O 2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

7 Solution 3) 76.2 % yield theoretical yield of CO 30.0 g O 2 x 1 mol O 2 x 2 mol CO x g CO g O 2 1 mol O 2 1 mol CO = 52.5 g CO (theoretical) percent yield 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO (theoretical)

8 Learning Check When N 2 and 5.00 g H 2 are mixed, the reaction produces 16.0 g NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 1) 31.3 % 2) 56.5 % 3) 80.0 %

9 Solution 2) 56.5 % N 2 (g) + 3H 2 (g) 2NH 3 (g) 5.00 g H 2 x 1 mol H 2 x 2 mol NH 3 x g NH g H 2 3 mol H 2 1 mol NH 3 = 28.2 g NH 3 (theoretical) Percent yield = 16.0 g NH 3 x 100 = 56.7 % 28.2 g NH 3