How much oxygen is needed to burn 13.5 g of aluminum?

The aluminum combustion reaction is described by the following chemical reaction equation:

4Al + 3O2 = 2Al2O3;

Let’s calculate the amount of substance contained in 13.5 grams of aluminum.

M Al = 27 grams / mol;

N Al = 13.5 / 27 = 0.5 mol;

To burn 0.5 mol of metal, 0.5 x 3/4 = 0.375 mol of oxygen is required.

Let’s calculate its volume.

1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V O2 = 0.375 x 22.4 = 8.4 liters;