What mass of salt is formed when aluminum reacts with 30 g of hydrochloric acid? How much gas was released during this?

Given:
m (HCl) = 30 g
To find:
m (AlCl3)
V (H2)
Decision:
6HCl + 2Al = 2AlCl3 + 3H2
n (HCl) = m / M = 30 g / 36.5 g / mol = 0.82 mol
n (HCl): n (AlCl3) = 3: 1
n (AlCl3) = 0.27 mol
m (AlCl3) = n * M = 0.27 mol * 133.5 g / mol = 36 g
n (HCl): n (H2) = 2: 1
n (H2) = 0.41 mol
V (H2) = n * Vm = 0.41 mol * 22.4 L / mol = 9.2 L
Answer: 36 g; 9.2 l