By burning 5.4 g of aluminum, alumina was obtained. Determine the amount of oxygen needed

By burning 5.4 g of aluminum, alumina was obtained. Determine the amount of oxygen needed to completely burn the 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 find the chemical amount of aluminum. For this purpose, we divide the weight of the existing substance by its molar weight.

M Al = 27 grams / mol;

N Al = 5.4 / 27 = 0.2 mol;

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

To do this, multiply the amount of oxygen by the volume of 1 mole of gas (filling volume 22.4 liters).

N O2 = 0.2 / 4 x 3 = 0.15 mol;

V O2 = 0.15 x 22.4 = 3.36 liters;