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

Atomic wt. of Fe = 56 and O = 16

Molecular mass of Fe₂O₃ = 2Fe + 3O

=(2 x 56) + (3 x 16)

= 112 + 48

= 160 g

Iron present in 80% of Fe₂O₃ = $\dfrac{112}{160}$ x 80

= 56 g

∴ Mass of iron in 100 g of iron ore = 56 g

Hence, mass of iron present in 10 kg (i.e., 10,000 g) of iron ore = $\dfrac{56}{100}$ x 10000

= 5600 g = 5.6 kg