4Fe(s)+3O2(g) →2Fe2O3(s) When 25.0 g of Fe reacts with 8.05 g of O2 , how many grams of Fe2O3 can form?

Answer 1

4 Fe + 3 O₂ → 2 Fe₂O₃

25g Fe x 1 mole Fe x 2 moles Fe₂O₃ x 156.69g Fe₂O_{3 =} 35.1 g of Fe₂O₃

55.85g Fe 4 mole Fe 1 mole Fe₂O₃

8.05g O₂ x 1 mole O₂ x 2 moles Fe₂O₃ x 156.69g Fe₂O_{3 =} 26.3g of Fe₂O₃

32g O₂ 3 moles O₂ 1 mole Fe₂O₃

The lowest amount tells you the maximum amount (theoretical yield) of the product that can be formed.

Therefore the amount of Fe₂O₃ that can be formed in this reaction is 26.3g.