How much air is required to interact with 5.4 g of aluminum containing 20% impurities?

The aluminum combustion reaction is described by the following chemical reaction equation:
4Al + 3O2 = 2Al2O3;
The weight of pure aluminum is 5.4 x 0.8 = 4.32 grams.
Let’s calculate the amount of substance contained in 4.32 grams of aluminum.
M Al = 27 grams / mol;
N Al = 4.32 / 27 = 0.16 mol;
To burn 0.16 mol of metal, 0.16 x 3/4 = 0.12 mol of oxygen is required.
Let’s calculate its volume.
1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.
V O2 = 0.12 x 22.4 = 2.688 liters;
Taking into account the oxygen content in the air at the level of 21%, the required air volume will be:
V air = 2.688 / 0.21 = 12.8 liters;