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:
m = 800 kg. X g -?
Fe2O3 + 2Al = 2Fe + Al2O3 – iron was obtained by the method of aluminothermy;
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.
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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