Question

Q1: If the theoretical yield of a reaction is 24.8 g and the actual yield is 18.5 g, what is percent yield?

Q2: When 11.5 gram of C are allowed to react with 114.5 g Cu2O, 87.4 g of Cu are obtained. Determine the limiting reactant, theoretical yield and percent yield.

Q3:This reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Determine limiting reactant, theoretical yield, and percent yield.

Answer 1

Q1: If the theoretical yield of a reaction is 24.8 g and the actual yield is 18.5 g, what is percent yield?

Percent yield = (actual yield/theoretical yield) * 100.

Percent yield = (18.5/24.8)*100 = 74.596%

Percent yield = 74.596%

Q2: When 11.5 gram of C are allowed to react with 114.5 g Cu₂O, 87.4 g of Cu are obtained. Determine the limiting reactant, theoretical yield and percent yield.

                   2Cu₂O + C ----> 4Cu + CO₂

First convert 11.5 gram of C, 114.5 g Cu2O and 87.4 g of Cu into moles

Moles of C = Wt/M.Wt = 11.5/12.01 = 0.957 moles

Moles of Cu₂O = Wt/M.Wt = 114.5/143.09 = 0.80 moles

Moles of Cu = Wt/M.Wt = 87.4/63.546 = 1.375 moles

                   2Cu₂O + C ----> 4Cu + CO₂

From equation 2 moles of Cu₂O reacts with one mole of C and generates 4 moles of Cu.

So for 2 moles of Cu2O, one mole of C is required.

So for given 0.80 moles of Cu2O, 0.40 moles of C is required. So 0.557 (0.957-0.40 = 0.557) moles of C will remain in the reaction. So limiting reagent in this reaction is Cu₂O.

Theoretical yield:

2 moles of Cu₂O reacts and generates 4 moles of Cu.

Then 0.80 moles Cu₂O generates 1.60 moles of Cu.

Convert 1.60 moles of Cu in to grams

Theretical yield (Weight)= moles x M.Wt = 1.60 x 63.546 = 101.67 gm

Theretical yield = 101.67 gm

Percent yield

Percent yield = (actual yield/theoretical yield) * 100.

Percent yield = (87.4/101.67)*100 = 85.96 %

Percent yield = 85.96%

Q3:This reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Determine limiting reactant, theoretical yield, and percent yield.

Fe₂O₃ + 3CO ==è 2Fe + 3CO₂

First convert 185 g of Fe2O3, 95.3 g of CO and 87.4 g of Fe into moles

Moles of Fe2O3 = Wt/M.Wt = 185/159.69 = 1.158 moles

Moles of CO = Wt/M.Wt = 95.3/28.01 = 3.40 moles

Moles of Fe = Wt/M.Wt = 87.4/55.845 = 1.565 moles

Fe₂O₃ + 3CO ==è 2Fe + 3CO₂

From equation one mole of Fe₂O₃ reacts with three moles of CO and generates 2 moles of Fe.

So for Three moles of CO, one mole of Fe₂O₃ is required.

So for given 3.40 moles of CO, 1.133 moles of Fe₂O₃ is required. So 0.024 (1.158-1.133 = 0.024) moles of Fe₂O₃ will remain in the reaction. So limiting reagent in this reaction is CO.

Theoretical yield:

one mole of Fe₂O₃ reacts and generates 2 moles of Fe.

Then 1.158 moles Fe₂O₃ generates 2.316 moles of Fe.

Convert 2.316 moles of Fe in to grams

Theretical yield (Weight)= moles x M.Wt = 2.316 x 55.845 = 129.33 gm

Theretical yield = 129.33 gm

Percent yield

Percent yield = (actual yield/theoretical yield) * 100.

Percent yield = (87.4/129.33)*100 = 67.579 %

Percent yield = 67.579%