Calculate what mass of steel containing 95% iron can be obtained from 11.6 tons of magnetic iron ore Fe3O4?

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

Fe3O4 + 2C = 3Fe + 2CO2.

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

11.6 t = 11,600,000 g.

n (Fe3O4) = m (Fe3O4) / M (Fe3O4) = 11600000 g / 232 g / mol = 50,000 mol.

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

n (Fe) = n (Fe3O4) * 3 = 50,000 mol * 3 = 150,000 mol.

m (Fe) = n (Fe) * M (Fe) = 150,000 mol * 56 g / mol = 8,400,000 g.

4. Find the mass of steel:

m (steel) = m (Fe) * 100% / ω (Fe) = 8400000 g * 100% / 95% = 8842105 g = 8.842 t.

Answer: m (steel) = 8.842 tons.