Combining 0.395 mol Fe2O3 with excess carbon produced 10.7 g Fe. Fe2O3+3C⟶2Fe+3CO. What is the actual yield of iron in moles? What is the theoretical yield and percent yield?

Answer 1

Fe₂O₃ + 3C ==> 2Fe + 3CO ... balanced equation

The actual yield of Fe (in moles) = 10.7 g Fe x 1 mol Fe / 55.85 g = 0.192 moles Fe

Theoretical yield = 0.395 mol Fe2O3 x 2 mol Fe / mol Fe2O3 x 160. g Fe2O3/mol = 126 g Fe2O3

Percent yield = actual yield / theoretical yield (x100%) = 10.7 g/126 g (x100%) = 8.49% yield