Calculate the mass and volume of oxygen required to completely burn 8.1 g of aluminum.

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

4Al + 3O2 = 2Al2O3;

Let’s calculate the chemical amount of aluminum. To do this, we divide the mass of the available substance by its molar weight.

M Al = 27 grams / mol;

N Al = 8.1 / 27 = 0.3 mol;

When burning aluminum for 1 mol of metal, it is necessary to take ¾ = 0.75 mol of oxygen. Let’s find its volume.

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

N O2 = 0.3 / 4 x 3 = 0.225 mol;

V O2 = 0.225 x 22.4 = 5.04 liters;