Calculate the mass of iron that can be obtained from 500 kg of Fe2O3 containing 2% impurities.

1. Let’s compose the equation of chemical reactions:

2Fe2O3 + 3C = 4Fe + 3CO2.

2. Find the chemical amount of iron oxide:

500 kg = 500000 g.

ω (Fe2O3) = 100% – ω (impurities) = 100% – 2% = 98%.

m (Fe2O3) = m (mixtures) * ω (Fe2O3) / 100% = 500,000 g * 98% / 100% = 490,000 g.

n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 490000 g / 160 g / mol = 3062.5 mol.

3. According to the reaction equation, we find the chemical amount of iron, and then its mass:

n (Fe) = n (Fe2O3) * 2 = 3062.5 mol * 2 = 6125 mol.

m (Fe) = n (Fe) * M (Fe) = 6125 mol * 56 g / mol = 343000 g = 343 kg.

Answer: m (Fe) = 343 kg.