4 Fe(s) + 3 O2(g)= 2 Fe2O3(s) What mass of O2 is required to oxidize 4543 g of iron to iron (III) oxide?

Molar mass of Fe = 55.85 g/mol
Molar mass of O₂ = 16.00 × 2 g/mol = 32.00 g/mol

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)
OR: Mole ratio Fe : O₂ = 4 : 3

No. of moles of Fe reacted = (4543 g) / (55.85 g/mol) = 81.34 mol
No. of moles of O₂ required = (81.34 mol) × (3/4) = 61.01 mol
Mass of O₂ required = (61.01 mol) × (32.00 g/mol) = 1952 g