Determine the volume of oxygen that interacts with aluminum if this produces 22.4 g of aluminum oxide.

The oxidation reaction of aluminum metal is described by the following chemical equation:

2Al + 3/2 O2 = Al2O3;

The aluminum oxide molecule contains 2 aluminum atoms.

Let’s calculate the amount of substance in 22.4 grams of metal oxide.

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

N Al2O3 = 22.4 / 102 = 0.2196 mol;

For the synthesis of such an amount of oxide, 0.2196 x 3/2 = 0.3294 mol of oxygen will be needed.

Let’s calculate its volume.

1 mole of ideal gas normally takes on a volume of 22.40 liters.

V O2 = 0.3294 x 22.40 = 7.38 liters;