How many liters of oxygen does it take to burn 54 g of aluminum?

Elemental aluminum reacts with oxygen gas 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 weight of the existing substance by its molar weight.

M Al = 27 grams / mol;

N Al = 54/27 = 2 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.

For this purpose, we multiply the amount of oxygen by the volume of 1 mole of gas (filling volume 22.4 liters).

N O2 = 2/4 x 3 = 1.5 mol;

V O2 = 1.5 x 22.4 = 33.6 liters;