Verified answer

If you have an actual yield of 12.2 g, and that is 66.8% of the theoretical yield, then, the theoretical yield would have been:

actual / theoretical X 100 = % yield

12.2 / Theoretical X 100 = 66.8

Theoretical = 18.3 grams Fe(s).

Now, work backwards from this to get to the starting mass of Fe2O3(s) needed:

18.3 g Fe X (1 mol/55.85 g) X (1 mol Fe2O3 / 2 mol Fe) X (159.7 g/mol Fe2O3) = 26.1 grams Fe2O3