During the oxidation of aluminum with oxygen, 800 g of Al2O3 were obtained.

During the oxidation of aluminum with oxygen, 800 g of Al2O3 were obtained. What volume of oxygen was required for oxidation?

Metallic aluminum reacts with oxygen to form alumina. The reaction is described by the following chemical reaction equation:

4Al + 3O2 = 2Al2O3;

Let’s determine the chemical amount of aluminum oxide. For this purpose, we divide the mass of the available substance by its molar weight.

M Al2O3 = 27 x 2 + 16 x 3 = 102 grams / mol;

N Al2O3 = 800/102 = 7.84 mol;

To get 1 mole of aluminum oxide, you need to take 3/2 mole of oxygen.

Let’s calculate the volume of oxygen.

To do this, multiply the amount of oxygen by the volume of 1 mole of gas (which is 22.4 liters).

N O2 = 7.84 x 3/2 = 11.76 mol;

V O2 = 11.76 x 22.4 = 263.42 liters;