Calculate the volume of oxygen required to burn 15 g of aluminum.

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 mass of the existing substance by its molar weight.

M Al = 27 grams / mol;

N Al = 15/27 = 0.555 mol;

When burning aluminum for 1 mol of metal, it is necessary to take ¾ = 0.75 mol of oxygen. Let’s find its volume.

To this end, multiply the amount of oxygen by the volume of 1 mole of gas (which is 22.4 liters).

N O2 = 0.555 / 4 x 3 = 0.41666 mol;

V O2 = 0.41666 x 22.4 = 9.333 liters;