Find the volume of hydrogen released during the mutual. aluminum with a solution of hydrochloric

Find the volume of hydrogen released during the mutual. aluminum with a solution of hydrochloric acid weighing 200 g and a mass fraction of a solute of 20%.

Given:
m solution (HCl) = 200 g
ω (HCl) = 20%

To find:
V (H2) -?

1) 2Al + 6HCl => 2AlCl3 + 3H2;
2) m (HCl) = ω * m solution / 100% = 20% * 200/100% = 40 g;
3) n (HCl) = m / M = 40 / 36.5 = 1.1 mol;
4) n (H2) = n (HCl) * 3/6 = 1.1 * 3/6 = 0.55 mol;
5) V (H2) = n * Vm = 0.55 * 22.4 = 12.3 liters.

Answer: The volume of H2 is 12.3 liters.