Question

When 1.0 mole of fe(s) reacts with excess o2(g) and 0.325 moles of fe2o3 are produced. what is the %yield of fe2o3?

Answer 1

The balanced reaction is:
4Fe + 3O2 --> 2Fe2O3
Stoichiometrically:
(1.0 mol Fe)(2 mol Fe2O3 / 4 mol Fe) = 0.50 mol Fe2O3
If the actual yield is only 0.325 mol Fe2O3, the % yield can be calculated by dividing actual by theoretical yield:
0.325 / 0.5 x 100% = 65% yield