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Theoretical yield vs. Actual yield. Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed,

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Presentation on theme: "Theoretical yield vs. Actual yield. Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed,"— Presentation transcript:

1 Theoretical yield vs. Actual yield

2 Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

3 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Based on: H 2 O = g O 2 Question: If only 35.2 g of O 2 were recovered, what was the percent yield? Hide 47.0 g H 2 O 125.3 Limiting/Excess Reactant Problem with % Yield Hide

4 2H 2 + O 2  2H 2 O What is the % yield of H 2 O if 58 g H 2 O are produced by combining 60 g O 2 and 7.0 g H 2 ? Hint: determine limiting reagent first 2 mol H 2 O 2 mol H 2 x # g H 2 O=7.0 g H 2 62.4 g= 18.02 g H 2 O 1 mol H 2 O x 1 mol H 2 2.02 g H 2 x 58 g H 2 O 62.4 g H 2 O = % yield = x 100%92.9%= actual theoretical x 100% 2 mol H 2 O 1 mol O 2 x # g H 2 O=60 g O 2 68 g= 18.02 g H 2 O 1 mol H 2 O x 1 mol O 2 32 g O 2 x

5 Once the limiting reactant of a reaction has been found, it can be used to calculate the theoretical yield The actual yield of a chemical reaction is less than the theoretical yield The actual yield must be determined by performing an experiment


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