How much oxygen must react with aluminum to produce 81.6 grams of aluminum oxide?

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

4Al + 3O2 = 2Al2O3;

Let’s find 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 = 81.6 / 102 = 0.8 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 (filling volume 22.4 liters).

N O2 = 0.8 x 3/2 = 1.2 mol;

V O2 = 1.2 x 22.4 = 26.88 liters;