To solve this problem, let us first find for the molar
mass of Fe2O3 and Fe3O4.
Fe = 55.85 g/mol and O = 16 g/mol
Therefore,
Fe2O3 = 159.7 g/mol
Fe3O4 = 231.55 g/mol
We are given that there are 29 g of Fe2O3, we calculate
for the amount of Fe from this in moles:
mol Fe = 29 g Fe2O3 (1 / 159.7 g/mol) (2 mol Fe / 1 mol
Fe2O3)
mol Fe = 0.363 mol
Converting this to Fe3O4:
mass Fe3O4 = 0.363 mol Fe (1 mol Fe3O4 / 3 mol Fe) (231.55
g/mol)
mass Fe3O4 = 28.03 g
Therefore there are 28.03g of Fe3O4 in the ore.
