Answer:
The actual yield of iron in moles = 0.338 moles
The theoretical yield of iron in moles = 0.700 moles iron
The percentage yield = 48.3 %
Explanation:
Step 1: Data given
Number of moles Fe2O3 = 0.350 moles
carbon = in excess
Mass of Fe = 18.9 grams
Step 2: The balanced equation
Fe2O3 + C ⟶ 2 Fe + 3 CO
Step 3: Calculate moles Fe
For 1 mol Fe2O3 we need 1 mol C to produce 2 moles Fe and 3 moles CO
For 0.350 moles Fe2O3 we'll have 2*0.350 = 0.700 moles Fe
Step 4: Calculate mass Fe
Mass Fe = moles Fe * molar mass Fe
Mass Fe = 0.700 moles * 55.845 g/mol
Mass Fe = 39.10 grams = the theoretical yield
Step 5: Calculate % yield
% yield = (actual yield / theoretical yield) * 100%
% yield = (18.9 grams / 39.10 grams )*100%
% yield = 48.3 %
Step 6: Calculate moles Fe actual yield
Moles Fe = 18.9 grams / 55.845 g/mol
Moles Fe = 0.338 moles
% yield in moles = (0.338 moles / 0.700 moles ) *100 %
% yield in moles = 48.2 %