What mass of aluminum is required to interact with iron (III) oxide weighing 120 kg?

What mass of aluminum is required to interact with iron (III) oxide weighing 120 kg? How much aluminum oxide is formed?

Given:
m (Fe2O3) = 120 kg = 120,000 g

Find:
m (Al) -?
n (Al2O3) -?

1) 2Al + Fe2O3 => 2Fe + Al2O3;
2) n (Fe2O3) = m / M = 120000/160 = 750 mol;
3) n (Al) = n (Fe2O3) * 2 = 750 * 2 = 1500 mol;
4) m (Al) = n * M = 1500 * 27 = 40500 g;
5) n (Al2O3) = n (Fe2O3) = 750 mol.

Answer: The mass of Al is 40500 g; the amount of Al2O3 substance is 750 mol.