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.
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?
1 answer