What mass of iron can be obtained by the reduction of 1 kg of magnetic iron ore containing

What mass of iron can be obtained by the reduction of 1 kg of magnetic iron ore containing 88.16% Fe3O4 with hydrogen?

Let’s find the mass of iron oxide (III) Fe3O4 in magnetic iron ore.

1000 g – 100%,

x – 88.16%.

X = (1000g × 88.16%): 100% = 881.6g.

Let’s find the amount of the substance Fe3O4.

n = m: M.

M (Fe3O4) = 232 g / mol.

n = 881.6 g: 232 g / mol = 3.8 mol.

Let’s find the quantitative ratios of substances.

Fe3O4 + 2H2 = 3Fe + 2H2O

There is 3 mol of Fe per 1 mol of Fe3O4.

The substances are in quantitative ratios of 1: 3.

The amount of Fe is 3 times more than Fe3O4.

n (Fe) = 3n (Fe3O4) = 3.8 mol × 3 = 11.4 mol.

Let us find the mass of Fe.

M (Fe) = 56 g / mol.

m = n × M.

m = 56 g / mol × 11.4 mol = 638.4 g.

Answer: 638.4 g.