How much oxygen is required for the reaction with m (Al) = 2.7?

The interaction of aluminum with oxygen is described by the following chemical reaction equation:

4Al + 3O2 = 2Al2O3;

Let’s determine the chemical amount of aluminum. To do this, divide its weight by the weight of 1 mole of the substance.

M Al = 27 grams / mol;

N Al = 2.7 / 27 = 0.1 mol;

With this amount of aluminum, 0.1 x 3/4 = 0.075 mol of oxygen will react.

Let’s calculate its volume.

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

V O2 = 0.075 x 22.4 = 1.68 liters;