What mass of iron can be obtained by hydrogen reduction of a mixture of ferum (2) oxide and ferum (3)

What mass of iron can be obtained by hydrogen reduction of a mixture of ferum (2) oxide and ferum (3) oxide weighing 110.4 grams, if the mass particles of oxides in the mixture are 13% and 87%, respectively?

1. Let us write down the equations of the reactions occurring during the reduction of iron from oxides:

FeO + H2 = Fe + H2O;

Fe2O3 + 3H2 = 2Fe + 3H2O;

2.Calculate the mass of oxides in the mixture:

m (FeO) = w (FeO) * m (mixtures) = 0.13 * 110.4 = 14.352 g;

m (Fe2O3) = w (Fe2O3) * m (mixture) = 0.87 * 110.4 = 96.048 g;

3.Calculate the chemical amounts of oxides:

n (FeO) = m (FeO): M (FeO);

M (FeO) = 56 + 16 = 72 g / mol;

n (FeO) = 14.352: 72 = 0.1993 mol;

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

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

n (Fe2O3) = 96.048: 160 = 0.6 mol;

4.Set the amounts of reduced iron in the first and second cases:

n1 (Fe) = n (FeO) = 0.1993 mol;

n2 (Fe) = n (Fe2O3) * 2 = 0.6 * 2 = 1.2 mol;

5. find the total amount of iron:

n (Fe) = n1 (Fe) + n2 (Fe) = 0.1993 + 1.2 = 1.3993 mol;

6.Calculate the mass of iron:

m (Fe) = n (Fe) * M (Fe) = 1.3993 * 56 = 78.36 g.

Answer: 78.36 g.