How many liters of oxygen will it take to burn 2.7 grams of aluminum?

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 = 2.7 / 27 = 0.1 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, we multiply the amount of oxygen by the volume of 1 mole of gas (which is 22.4 liters).
N O2 = 0.1 / 4 x 3 = 0.075 mol;
V O2 = 0.075 x 22.4 = 1.68 liters;