How many grams of salt is formed by dissolving 13.5 grams of aluminum in hydrochloric acid?

Let’s write the reaction equation:

2Al + 6HCl = 2AlCl3 + 3H2 ↑

Let’s find the amount of aluminum substance:

v (Al) = m (Al) / M (Al) = 13.5 / 27 = 0.5 (mol).

According to the reaction equation, 2 mol of Al is formed from 2 mol of AlCl3, therefore:

v (AlCl3) = v (Al) = 0.5 (mol).

Thus, the mass of the formed aluminum chloride salt is:

m (AlCl3) = v (AlCl3) * M (AlCl3) = 0.5 * 133.5 = 66.75 (g).

Answer: 66.75 g.