Question

Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.

Answer 1

In Fe₂O₃ , Fe = 56 and O = 16

Molecular mass of Fe₂O₃ = 2 × 56 + 3 ×16 = 160 g

Iron present in 80% of Fe₂O₃ = `112/160 xx 80 = 56`g

So, mass of iron in 100 g of ore = 56 g

mass of Fe in 10000 g of ore = `56 xx 10000/100`

= 5.6 kg