# The reaction involved 0.2 mol of aluminum hydroxide and 0.5 mol of hydrochloric acid.

The reaction involved 0.2 mol of aluminum hydroxide and 0.5 mol of hydrochloric acid. Determine the amount of the formed salt.

n (Al (OH) 3) – 0.2 mol.

n (HCl) – 0.5 mol.

Determine the amount of salt formed. We write down the solution.

First, you need to write down the reaction equations.

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

Next, we determine which substance is in excess.

X mol – 0.5 mol.

1 mol – 3 mol.

X = 1 * 0.5: 3 = 0.167 mol.

Therefore, aluminum hydroxide is in excess. We calculate using hydrochloric acid.

0.5 mol – x mol.

3 mol – 1 mol.

X = 0.5 * 1: 3 = 0.167 mol of salt.

Answer: 0.167 mol of aluminum chloride is formed.

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