Determine the mass of aluminum and the volume of chlorine required to obtain aluminum chlorine weighing 80.1 grams.

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 80.1 grams of aluminum chloride.

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

N AlCl3 = 80.1 / 133.5 = 0.6 mol;

To teach such an amount of salt, 0.6 mol of metal and 0.9 mol of chlorine are required.

Let’s calculate the mass of aluminum.

m Al = 27 x 0.6 = 16.2 grams;

Let’s calculate the volume of chlorine.

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

N Cl2 = 0.9 x 22.40 = 20.16 liters;