Calculate the mass and volume of oxygen required to burn 30 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 30 grams of aluminum.

M Al = 27 grams / mol;

N Al = 30/27 = 1.111 mol;

To burn 1.111 mol of metal, 1.111 x 3/4 = 0.833 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 = 0.833 x 22.4 = 18.659 liters;

We calculate the weight of 0.833 mol of oxygen.

M O2 = 16 x 2 = 32 grams / mol;

m O2 = 32 x 0.833 = 26.656 grams;