What mass of iron(III) oxide must be used to produce 87.90 g of iron?

Method 1 :

Molar mass of Fe₂O₃ = (55.85×2 + 16.00×3) g/mol = 159.7 g/mol
Molar mass of Fe = 55.85 g/mol

Fe₂O₃ + other reactant(s) → 2Fe + other products
Mole ratio Fe₂O₃ : Fe = 1 : 2

No. of moles of Fe produced = (87.90 g) / (55.85 g/mol) = 1.574 mol
No. of moles of Fe₂O₃ used = (1.574 mol) × (1/2) = 0.787 mol
Mass of Fe₂O₃ used = (0.787 mol) × (159.7 g/mol) = 125.7 g


====
Method 2 :

Mass of 1 mole of Fe₂O₃ = (55.85×2 + 16.00×3) g = 159.7 g
Mass of Fe in 1 mole of Fe₂O₃ = 55.85×2 g = 111.7 g
Mole fraction of Fe in Fe₂O₃ = 111.7/159.7

Mass of Fe₂O₃ used = (87.90 g Fe) × [(159.7 g Fe₂O₃) / (111.7 g Fe)] = 125.7 g Fe₂O₃