What mass of aluminum chloride is formed by the interaction of hydrochloric acid weighing 7.3 g with aluminum?

1. Let’s write the equation of interaction of hydrochloric acid with aluminum:

2Al + 6HCl = 2AlCl3 + 3H2 ↑;

2.Calculate the chemical amount of hydrogen chloride:

n (HCl) = m (HCl): M (HCl) = 7.3: 36.5 = 0.2 mol;

3. Determine the amount of aluminum chloride from the proportion:

6 mol HCl – 2 mol AlCl3;

0.2 mol HCl – x mol AlCl3;

x = 0.2 * 2: 6 = 0.0667 mol;

4.Calculate the mass of the resulting chloride:

m (AlCl3) = n (AlCl3) * M (AlCl3);

M (AlCl3) = 27 + 35.5 * 3 = 133.5 g / mol;

m (AlCl3) = 0.0667 * 133.5 = 8.9 g.

Answer: 8.9 g