What mass of aluminum is formed during the decomposition of 145.7 g of aluminum oxide containing 30% impurities?

We implement the solution:
1. We compose the equation according to the condition of the problem:
2Al2O3 = 4Al + 3O2 – decomposition, aluminum is obtained;
2. Let’s make calculations using the formulas:
M (Al2O3) = 101.8 g / mol.
M (Al) = 26.9 g / mol.
m (Al2O3) = 145.7 * (1 – 0.3) = 101.9 g (mass without impurities).
Y (Al2O3) = m / M = 101.9 / 101.8 = 1 mol.
3. Proportion:
1 mol (Al2O3) – X mol (Al);
-2 mol -4 mol from here, X mol (Al) = 1 * 4/2 = 2 mol.
4. Find the mass of the product:
m (Al) = Y * M = 2 * 26.9 = 53.8 g.
Answer: We obtained aluminum with a mass of 53.8 g.