Question

Combining 0.339 mol Fe₂O₃ with excess carbon produced 11.7 g Fe.

Fe₂0₃ + 3C → 2Fe + 3CO
What is the actual yield of iron in moles? What is the theoretical yield of iron in moles? What is the percent yield?

Answer 1

The actual yield of iron in moles is 0.209 mol. The theoretical yield of iron in moles is 0.339 mol. The percent yield is 61.65%.

Step-by-step explanation:

The actual yield of iron in moles can be calculated by dividing the given mass of iron by the molar mass of iron (55.85 g/mol). In this case, the actual yield is 11.7 g / 55.85 g/mol = 0.209 mol.

The theoretical yield of iron in moles can be calculated using the stoichiometric mole ratio from the balanced chemical equation. In this case, the mole ratio between Fe2O3 and Fe is 2:2, which means that for every mole of Fe2O3, 2 moles of Fe are produced. Therefore, the theoretical yield is equal to the moles of Fe2O3 used, which is 0.339 mol.

The percent yield can be calculated by dividing the actual yield by the theoretical yield, and then multiplying by 100. In this case, the percent yield is (0.209 mol / 0.339 mol) * 100 = 61.65%.