4Fe(s)+3O2(g) →2Fe2O3(s) If 74.9 g of Fe and 67.8 g of O2 are mixed, and 85.0 g of Fe2O3 is actually obtained, what is the percent yield of Fe2O3 for the reaction?

Answer 1

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ... balanced equation

To find the theoretical yield, we first need to determine which reactant is limiting. An easy way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less is limiting.

For Fe: 74.9 g Fe x 1 mol Fe / 55.85 g = 1.34 mol Fe (÷4->0.335) LIMITING

For O₂: 67.8 g O₂ x 1 mol O₂ / 32 g = 2.12 mols O₂ (÷3->0.71)

Now use MOLES of limiting reactant (1.34 mol Fe) to find the theoretical yield.

1.34 mol Fe x 2 mol Fe₂O₃ / 4 mol Fe x 159.7 g Fe₂O₃ / mol = 107 g Fe₂O₃

% yield = actual yield / theoretical yield (x100%) = 85.0 g / 107 g (x100%)

% yield = 79.4% yield