200 g of Al2O3 was treated with 100 g of HCl. Find the mass and amount of salt.

Let’s implement the solution:

In accordance with the condition, we will write the data:
Al2O3 + 6HCl = 2AlCl3 + 3H2O – ion exchange, aluminum chloride was obtained.

We make calculations using the formulas:
M (Al2O3) = 101.8 g / mol;

M (HCl) = 36.5 g / mol;

M (AlCl3) = 133.4 g / mol.

Let’s determine the amount of starting materials:
Y (Al2O3) = m / M = 200 / 101.8 = 1.96 mol (deficient substance);

Y (HCl) = m / M = 100 / 36.5 = 2.7 mol (substance in excess).

Calculations are carried out for the substance in deficiency.

Proportion:
1.96 mol (Al2O3) – X mol (AlCl3);

1 mol -2 mol hence, X mol (AlCl3) = 1.96 * 2/1 = 3.9 mol.

Find the mass of the product:
m (AlCl3) = Y * M = 3.9 * 133.4 = 522.9 g

Answer: the amount of aluminum chloride is 3.9 mol, the mass is 522.9 g