From 320 g of iron (III) oxide, aluminum reduced 112 g of iron. Determine the product yield

From 320 g of iron (III) oxide, aluminum reduced 112 g of iron. Determine the product yield as a percentage of what is theoretically possible.

Let’s write the reaction equation:

Fe2O3 + 2Al → Al2O3 + 2Fe

Find the amount of iron (III) oxide substance:

n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 320 g / 160 g / mol = 2 mol.

Let’s find the amount of iron substance (practical):

n (Fe) = 2 * n (Fe2O3) = 2 * 2 mol = 4 mol.

mpract. = n (Fe) * M (Fe) = 4 mol * 56 g / mol = 224 g.

Let’s use the formula for the mass fraction of the product yield:

n = (mpract. * 100%) / mtheoret. = (112 g * 100%) / 224 g = 50%

Answer: 50%.