How much air is required to interact with 6 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 6 x 0.8 = 4.8 grams.

Let’s calculate the amount of substance contained in 4.8 grams of aluminum.

M Al = 27 grams / mol;

N Al = 4.8 / 27 = 0.178 mol;

To burn 0.178 mol of metal, 0.178 x 3/4 = 0.134 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 = 0.134 x 22.4 = 3 liters;

Taking into account the oxygen content in the air at the level of 21%, the required air volume will be:

V air = 3 / 0.21 = 14.29 liters;