Calculate the amount of oxygen that will be required to react with 54 g of aluminum.

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

4Al + 3O2 = 2Al2O3;

Let’s calculate the amount of a substance contained in 54 grams of aluminum.

M Al = 27 grams / mol;

N Al = 54/27 = 2 mol;

To burn 2 mol of metal, 2 x 3/4 = 1.5 mol of oxygen will be required.

Let’s calculate its volume.

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

V O2 = 1.5 x 22.4 = 33.6 liters;