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Limiting Reactants (Reagents) and Percent Yield. Calculations need to be based on the limiting reactant. Example 1: Suppose a box contains 87 bolts, 110.

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Presentation on theme: "Limiting Reactants (Reagents) and Percent Yield. Calculations need to be based on the limiting reactant. Example 1: Suppose a box contains 87 bolts, 110."— Presentation transcript:

1 Limiting Reactants (Reagents) and Percent Yield

2 Calculations need to be based on the limiting reactant. Example 1: Suppose a box contains 87 bolts, 110 washers and 99 nails. How many sets of 1 bolt, 2 washers and 1 nail can you use to create? What is the limiting factor? 55 sets; washers limit the amount

3 Calculations need to be based on the limiting reactant. Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + O 2  CO 2 + SO 2 Start by balancing the equation…

4 Calculations need to be based on the limiting reactant. Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO 2 Now solve the problem…

5 Calculations need to be based on the limiting reactant. Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO 2 Which reactant is LIMITING?

6 Calculations need to be based on the limiting reactant. Example 2: What is the mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO 2 O 2 limits the amount of SO 2 that can be produced. CS 2 is in excess.

7 Calculations need to be based on the limiting reactant. Example 2: What is the mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen? CS 2 + 3O 2  CO 2 + 2SO g of SO 2 can be produced in this reaction.

8 Calculations need to be based on the limiting reactant. Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? CH 4 + O 2  CO 2 + H 2 O Start by balancing the equation…

9 Calculations need to be based on the limiting reactant. Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O Now solve the problem…

10 Calculations need to be based on the limiting reactant. Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O Which reactant is LIMITING?

11 Calculations need to be based on the limiting reactant. Example 3: What mass of CO 2 could be formed by the reaction of 8.0 g CH 4 with 48 g O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O CH 4 limits the amount of CO 2 that can be produced. O 2 is in excess.

12 Many chemical reactions do not go to completion (reactants are not completely converted to products). Percent Yield: indicates what percentage of a desired product is obtained.

13 So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%. The THEORETICAL YIELD is the yield calculated from the balance equation. The ACTUAL YIELD is the amount “actually” obtained in an experiment.

14 Look back at Example 2. We found that 134 g of SO 2 could be formed from the reactants. In an experiment, you formed 130 g of SO 2. What is your percent yield?

15 Example: A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this? CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O

16 Example: A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this?


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