The Fe is our limiting reactant here. First, convert the mass of Fe given to moles of Fe by dividing mass by the molar mass:
(51.3 g Fe)/(55.845 g/mol) = 0.9186 mol Fe.
According to the balanced equation, 2 moles of Fe2O3 are produced for every 4 moles of Fe that are consumed. In other words, half as many moles of Fe2O3 will be produced as there are Fe that react:
(0.9186 mol Fe)(2 mol Fe2O3/4 mol Fe) = 0.4593 mol Fe2O3.
Finally, convert the moles of Fe2O3 back to grams by multiplying by the molar mass of Fe2O3:
(0.4593 mol Fe2O3)(159.6882 g/mol) = 73.3 g Fe2O3.
So, 73.3 grams of Fe2O3 are formed.
