# Calculate the mass of iron that is formed during the reduction of 800 kg of iron oxide (3)

Calculate the mass of iron that is formed during the reduction of 800 kg of iron oxide (3) containing 20% impurities with aluminum.

Let’s execute the solution:
1. According to the condition of the problem, we compose the equation:
20% impurities;
m = 800 kg. X g -?
Fe2O3 + 2Al = 2Fe + Al2O3 – iron was obtained by the method of aluminothermy;
2. Calculations:
m (Fe2O3) = 800,000 g;
M (Fe2O3) = 159.6 g / mol;
M (Fe) = 55.8 g / mol.
3. Determine the mass of the original substance without impurities, as well as its amount:
800,000 (1 – 0.20) = 640,000 g;
Y (Fe2O3) = m / M = 640,000 / 159 = 4010 mol.
4. Proportion:
4010 mol (Fe2O3) – X mol (Fe);
-1 mol -2 mol from here, X mol (Fe) = 4010 * 2/1 = 8020 mol.
5. Find the mass of Fe:
m (Fe) = Y * M = 8020 * 55.8 = 447516 g. = 447 kg 516 g.
Answer: Iron was obtained with a mass of 447 kg. 516 BC

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