Calculate the mass of aluminum oxide that is formed as a result of the combustion of 3 grams

Calculate the mass of aluminum oxide that is formed as a result of the combustion of 3 grams of aluminum powder containing 10% impurities. How much oxygen was required for this?

Metallic 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 mass of the existing substance by its molar weight.

M Al = 27 grams / mol;

N Al = 3 x 0.9 / 27 = 0.1 mol;

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

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 = 0.1 / 2 = 0.05 mol;

m Al2O3 = 0.05 x 102 = 5.1 grams;

The required amount of oxygen will be 0.1 x 3/4 = 0.075 mol;