Determine the mass of the oxide that was obtained by the interaction of 11.2 liters of oxygen with aluminum.

Reaction equation: 4Al + 3O2 = 2Al2O3
Let’s find the amount of oxygen substance n = V / Vm = 11.2 l / 22.4 l / mol = 0.5 mol. (Vm – molar volume, constant for all gases under normal conditions). According to the reaction, 2 mol of oxide is formed from 3 mol of oxygen, then 0.5 / 3 = n (Al2O3) / 2.n (Al2O3) = 0.5 * 2/3 = 0.33 mol. Then the mass of the oxide is m = n * M = 0.33 mol * (27 * 2 + 16 * 3) g / mol = 33.66 g (M is the molar mass, determined by the periodic table).
Answer: 33.66 g of oxide was obtained.