Question

Iron rusts according to the balanced equation: 4Fe(s) +3O2(g) -> 2Fe2O3(s). Since the molecular weight of iron (Fe) is 56, 4 moles weigh 4 x 56 = 224 g. The molecular weight of Fe2O3 is (2 atoms x 56) + (3 atoms x 16) which is 160 g, so 2 moles of Fe2O3 weigh 2 x 160 = 320 g. Thus, 224 g of Fe produces 320 g of Fe2O3. Then, 500 g of Fe produces how many g of Fe2O3?

Answer 1

Answer:

714.28 g of Fe₂O₃

Explanation:

4 moles of Fe = 224g

3 moles of O₂ = 96g

Ratio by mass,

⇒ 224 : 96

⇒ 7 : 3

⇒ 500 g of Fe : 1500 ÷ 7 g of O₂

⇒ 500 g of Fe : 214.28 g of O₂

So, Total mass ⇒ 714.28 g of Fe₂O₃