How much oxygen is required to interact with zinc weighing 400 kg, containing 30% impurities?

Based on the data, we write down the equation of the process:
2Zn + O2 = 2ZnO – compounds, zinc oxide was formed;

Calculations:
M (Zn) = 65.3 g / mol;

M (O2) = 32 g / mol;

m (Zn) = 400 kg = 400,000 g;

m (Zn) = 400,000 * (1 – 0.30) = 280,000 g (mass without impurities).

Determine the amount of the original substance:
Y (Zn) = m / M = 280,000 / 65.3 = 4287.9 mol.

Proportion:
4287.9 mol (Zn) – X mol (O2);

-2 mol -1 mol from here, X mol (O2) = 4287.9 * 1/2 = 2144 mol.

We find the volume of O2:
V (O2) = 2144 * 22.4 = 48025.6 L

Answer: you need oxygen with a volume of 48025.6 liters