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

It is given that Iron ore is 80% pure. So 80% of 10Kg = 8Kg of Fe2O3
Number of moles of Fe2O3= mass/molar mass
= 8000/159.7 = 50.09 moles
One mole of Fe2O3 contains 2 moles of Fe
So number of moles of Fe= 50.09 x 2 = 100.18 moles
Mass of Fe = moles x atomic mass = 100.18 x 55.85 = 5595.05g = 5.595 kg.