How many grams of aluminum oxide is formed when 15 g aluminum interacts with oxygen?

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 = 15/27 = 0.555 mol;

When oxidizing this amount of aluminum, 2 times less amount of aluminum oxide will be obtained. Let’s define its mass.

For this purpose, we multiply the amount of the substance obtained by its molar weight.

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

N Al2O3 = 0.555 / 2 = 0.2777 mol;

m Al2O3 = 0.2777 x 102 = 28.333 grams;