Determine what amount of chloride 3 (mol) was formed by the interaction of amphoteric

Determine what amount of chloride 3 (mol) was formed by the interaction of amphoteric hydroxide aluminum 3 with hydrochloric acid weighing 180 grams (mass fraction of hydrochloric acid 30%).

The reaction of interaction of aluminum hydroxide with hydrochloric acid is described by the following chemical equation:

Al (OH) 3 + 3HCl = AlCl3 + 3H2O;

When 1 molecule of alkali is dissolved, 1 molecule of aluminum chloride is formed. This requires 3 molecules of hydrochloric acid.

Let’s calculate the chemical amount of a substance in 180 grams of 30% acid.

M HCl = 1 + 35.5 = 36.5 grams / mol;

N HCl = 180 x 0.3 / 36.5 = 1.48 mol;

With this amount of acid, 1.48 / 3 = 0.49 mol of aluminum chloride can be synthesized.