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?

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