Combining 0.302 mol Fe2O3 with excess carbon produced 10.4 g Fe.
Fe2O3+3C⟶2Fe+3CO
What is the actual yield of iron in moles?
What is the theoretical yield of iron in moles?
What is the percent yield?
actual yield in grams is 10.4 g. moles = grams/atomic mass Fe = 10.4/55.85 = 0.186 moles Fe
Theoretical yield calculated as follows:
Fe2O3 +3C⟶2Fe + 3CO
moles Fe2O3 = 0.302. Convert to mols Fe formed. That's
0.302 mols Fe2O3 x (2 moles Fe/1 mol Fe2O3) = 0.302 x 2 = 0.604moles Fe.
%yield = (actual yield/theor yield)*100 = (0.186/0.604)*100 = 30.8%
Of if done by grams = 10.4 g Fe actual yield
Theoretical yield = 0.604 x 55.85 = 33.73 g
%yield =
(10.4/33.73)*100 = 30.8% so percent yield with grams or with moles gives the same answer but with using different numbers; i.e., either grams or moles.