Calculate what mass of magnetite Fe3O4, containing 10% impurities, is required to obtain iron weighing 4 tons.

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

Fe3O4 + 4H2 = 3Fe + 4H2O.

2. Let’s find the chemical amount of the obtained iron:

4 t = 4,000,000 g.

n (Fe) = m (Fe) / M (Fe) = 4,000,000 g / 56 g / mol = 71428.57 mol.

3. Using the reaction equation, we find the amount of magnetite, and then its mass:

n (Fe3O4) = n (Fe) / 3 = 71428.57 mol / 3 = 23809.5 mol.

m (Fe3O4) = n (Fe3O4) * M (Fe3O4) = 23809.5 mol * 232 g / mol = 5523809.5 g.

4. Let’s find the mass of technical magnetite:

ω (Fe3O4) = 100% – ω (impurities) = 100% – 10% = 90%.

mtechn (Fe3O4) = m (Fe3O4) * 100% / ω (Fe3O4) = 5523809.5 g * 100% / 90% = 6137566.1 g = 6.138 t.

Answer: mtechn (Fe3O4) = 6.138 t.