For the following reaction: Fe2O3 + 3CO - - > 2Fe + 3CO2 how many grams of Fe2O3 is needed to produce 111 g of Fe?

You need 158.70 grams of Fe2O3 to produce 111 grams of Fe. This is calculated by using the molar masses and stoichiometric relationship of the two compounds.

Solution:

MM Fe = 55.845 g/mol

MM Fe2O3 = 159.69 g/mol

Fe: Fe2O3 = 2 mol:1 mol

11 g FE (1 mol Fe/55.845 g Fe) (1 mol Fe2O3/2 mol Fe) (159.69 g Fe2O3 / 1 mole Fe2O3) = 158.70 grams Fe2O3