500
This would be the percent yield if 11 g of iron metal oxidized with excess oxygen to form 12 grams of iron (III) oxide.
What is 75%
Fe=56 g/mol
4 Fe + 3 O2 --> 2 Fe2O3
11/56=0.2, 0.2mol Fe x (2 mol Fe2O3/ 4 mol Fe)= 0.1
0.1 mol Fe2O3 x 160 g/ mol = 16 (theoretical yield)
(2 x 56 + 3 x 16 )= 48 + 112 = 160 g/ mol
12/16=0.75, 0.75 x 100% = 75 %