80 g of iron (III) oxide is reduced with carbon monoxide to iron. Find the mass of the obtained iron

80 g of iron (III) oxide is reduced with carbon monoxide to iron. Find the mass of the obtained iron if the product yield is 90%.

Let’s write the reaction equation:

Fe2O3 + 3CO → 2Fe + 3CO2 (iron, carbon dioxide).

1) Calculate the amount of iron (III) oxide substance:

n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 80 g / 160 g / mol = 0.5 mol.

2) Let’s calculate the amount of iron substance:

n (Fe) = n (Fe2O3) = 0.5 mol.

3) Calculate the theoretical mass of iron:

mtheor. (Fe) = n (Fe) * M (Fe) = 0.5 mol * 56 g / mol = 28 g.

4) Let’s find the practical mass of the obtained iron:

mpract. (Fe) = (mtheoret. (Fe) * n (Fe)) / 100% = (28 g * 90%) / 100% = 25.2 g.

Answer: 25.2 grams.