Determine the amount of chlorine required to obtain 634 g of aluminum chloride Reaction scheme: Al + Cl2 = AlCl3

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, it is necessary to take 1 mol of metal and 1.5 mol of gaseous chlorine.

Let’s calculate the chemical amount of a substance that is in 634 grams of aluminum chloride.

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

N AlCl3 = 634 / 133.5 = 4.75 mol;

To teach this amount of salt, you need to take 4.75 x 3/2 = 7.125 mol of chlorine.

Find the volume of chlorine gas.

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

N Cl2 = 7.125 x 22.40 = 159.6 liters;