What is the mass of aluminum oxide formed by the interaction of 35 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 = 35/27 = 1.2963 mol;

When burning this amount of aluminum, 2 times less amount of aluminum oxide will be synthesized. Let’s find its weight.

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

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

N Al2O3 = 1.2963 / 2 = 0.648 mol;

m Al2O3 = 0.648 x 102 = 66.1 grams;