Find the mass of iron (III) oxide required to obtain pure iron weighing 33.6 g. Find V hydrogen required for the reaction.

1. Let’s write down the reaction of reduction of iron from its oxide by hydrogen:

Fe2O3 + 3H2 = 2Fe + 3H2O;

2.find the chemical amount of iron:

n (Fe) = m (Fe): M (Fe) = 33.6: 56 = 0.6 mol;

3.determine the chemical amount of iron (III) oxide and hydrogen:

n (Fe2O3) = n (Fe): 2 = 0.6: 2 = 0.3 mol;

n (H2) = n (Fe) * 3: 2 = 0.6 * 3: 2 = 0.9 mol;

4.Calculate the mass of iron (III) oxide:

m (Fe2O3) = n (Fe2O3) * M (Fe2O3);

M (Fe2O3) = 56 * 2 + 3 * 16 = 160 g / mol;

m (Fe2O3) = 0.3 * 160 = 48 g;

5. find the volume of hydrogen:

V (H2) = n (H2) * Vm = 0.9 * 22.4 = 20.16 dm3.

Answer: 48 g Fe2O3, 20.16 dm3 H2.