When aluminum (Al) interacts with oxygen (O2), aluminum oxide weighing 153 g

When aluminum (Al) interacts with oxygen (O2), aluminum oxide weighing 153 g was formed. Determine what volume of oxygen entered into the reaction.

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 oxide. For this purpose, we divide the weight of the existing substance by its molar weight.

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

N Al2O3 = 153/102 = 1.5 mol;

To get 1 mole of aluminum oxide, you need to take 3/2 mole of oxygen.

Let’s calculate the volume of oxygen.

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

N O2 = 1.5 x 3/2 = 2.25 mol;

V O2 = 2.25 x 22.4 = 50.4 liters;