Determine the amount of chlorine required to obtain 534 aluminum chloride.

Aluminum chloride has the following chemical formula AlCl3.

The reaction of its synthesis is described by the following equation:

Al + 3/2 Cl2 = AlCl3;

To obtain 1 mol of aluminum chloride, you need to take 1 mol of metal and 1.5 mol of gaseous chlorine.

Let’s calculate the chemical amount of a substance in 534 grams of aluminum chloride.

M AlCl3 = 27 + 35.5 x 3 = 133.5 grams / mol;

N AlCl3 = 534 / 133.5 = 4 mol;

To teach this amount of salt, you need 4 x 3/2 = 6 mol of chlorine.

Let’s calculate the volume of chlorine.

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

N Cl2 = 6 x 22.40 = 134.4 liters;