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.
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