How many moles of oxygen is required for oxidation 140 gr. 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. For this purpose, we divide the mass of the available substance by its molar weight.

M Al = 27 grams / mol;

N Al = 140/27 = 5.185 mol;

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

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

N O2 = 5.185 / 4 x 3 = 3.888 mol;

V O2 = 3.888 x 22.4 = 87.111 liters;