What mass of iron can be obtained as a result of the reduction of 16 kg of Fe2O3 if the reaction

What mass of iron can be obtained as a result of the reduction of 16 kg of Fe2O3 if the reaction yield from the theoretical is 90%.

1. Make up the equation:
M = 16 kg. X g -?
Fe2O3 + 3H2 = 2Fe + 3H2O – OBP, iron obtained;
2. We make calculations by the formulas:
M (Fe2O3) = 159.6 g / mol;
M (Fe) = 55.8 g / mol.
3. Determine the number of moles of iron oxide:
Y (Fe2O3) = m / M = 16000 / 159.6 = 100.25 mol.
4. According to the equation, we compose the proportion:
100.25 mol (Fe2O3) – X mol (Fe);
-1 mol – 2 mol from here, X mol (Fe) = 100.25 * 2/1 = 200.5 mol.
5. Find the theoretical mass of the reaction product:
m (Fe) = Y * M = 200.5 * 55.8 = 11187.9 g.
the practical mass is calculated taking into account the product yield:
W = m (practical) / m (theoretical) * 100;
m (practical) = 0.90 * 11187.9 = 10069 g = 10 kg 69 g.
Answer: in the process of recovery, iron was obtained with a mass of 10 kg. 69 g