Determine the mass of aluminum oxide obtained by the interaction of 13.5 g of aluminum with oxygen.

The reaction of aluminum with oxygen proceeds according to the following chemical formula:

4Al + 3O2 = 2Al2O3;

From two moles of aluminum, one mole of aluminum oxide is formed.

Let’s find the amount of a substance contained in 13.5 grams of aluminum:

N Al = 13.5 / 27 = 0.5 mol;

The amount of aluminum oxide substance will be:

N Al2O3 = N Al / 2 = 0.5 / 2 = 0.25 mol;

Let’s find the mass of this amount of aluminum oxide:

M Al2O3 = 27 x 2 + 16 x 3 = 102 grams / mol;

m Al2O3 = 0.25 x 102 = 25.5 grams;