Determine the mass of aluminum oxide that will result from the interaction of 140 g of aluminum 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 calculate the chemical amount of aluminum. To do this, we divide the mass of the existing substance by its molar weight.

M Al = 27 grams / mol;

N Al = 140/27 = 5.185 mol;

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

To this end, we multiply the amount of the substance obtained by its molar weight.

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

N Al2 O3 = 5.185 / 2 = 2.59 mol;

m Al2O3 = 2.59 x 102 = 264.4 grams;