Theoretical yield vs. Actual yield

1. Theoretical yield vs.Actual yield

2. 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

3. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide Hide 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Question: If only 35.2 g of O2 were recovered, what was the percent yield?

4. 1 mol O2 actual 1 mol H2 2 mol H2O 2 mol H2O 58 g H2O x x x x = 32 g O2 2.02 g H2 theoretical 1 mol O2 2 mol H2 62.4 g H2O 18.02 g H2O 18.02 g H2O x x 1 mol H2O 1 mol H2O = = 92.9% 68 g = 62.4 g 2H2 + O2 2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first # g H2O= 60 g O2 # g H2O= 7.0 g H2 % yield = x 100% x 100%

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