For the reaction, we took aluminum with a mass of 5.4 g and hydrochloric acid with a mass of 25.55 g

For the reaction, we took aluminum with a mass of 5.4 g and hydrochloric acid with a mass of 25.55 g, calculate the mass of the salt.

Given: m (Al) = 5.4 g; m (HCl) = 25.55 g
Decision:
2Al + 6HCl = 2AlCl3 + 3H2 – reaction of aluminum with hydrochloric acid;
n (Al) = m / M = 5.4 g: 27 g / mol = 0.2 mol – amount of aluminum;
n (HCl) = m / M = 25.55 g: 36.5 g / mol = 0.7 mol – amount of hydrochloric acid;
n (Al) / n (HCl) = 2/6 = 1/3 – the ratio of the number of substances in the reaction;
0.2 / 0.7 ≠ 1/3 => HCl in excess;
n (Al) = n (AlCl3) = 0.2 mol;
m (AlCl3) = nM = 0.2 mol * 133.5 g / mol = 26.7 g – mass of salt
Answer: 26.7 g.