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 transcript:

Theoretical yield vs. Actual yield

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

4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction contains 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) g47.0 g ? g Based on: KO 2 = g O g KO 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 Limiting/Excess Reactant Problem with % Yield Hide

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 g= g H 2 O 1 mol H 2 O x 1 mol H 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= g H 2 O 1 mol H 2 O x 1 mol O 2 32 g O 2 x

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