How much of an alumina substance can be obtained if 67.5 g of alumina have entered into a reaction with oxygen?

Metallic aluminum reacts with oxygen to form alumina. The reaction is described by the following chemical reaction equation:

4Al + 3O2 = 2Al2O3;

Let’s determine the chemical amount of aluminum. To this end, we divide the weight of the substance present by its molar weight.

M Al = 27 grams / mol;

N Al = 67.5 / 27 = 2.5 mol;

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

To do this, we multiply the amount of the resulting substance by its molar weight.

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

N Al2O3 = 2.5 / 2 = 1.25 mol;

m Al2O3 = 1.25 x 102 = 127.5 grams;