Calculate the volume of gas that will be released when aluminum interacts with a 40% solution of potassium

Calculate the volume of gas that will be released when aluminum interacts with a 40% solution of potassium hydroxide in a volume of 100 ml with a density of 1.25 ml.

Given:
ω (KOH) = 40%
V solution (KOH) = 100 ml
ρ solution (KOH) = 1.25 g / ml

To find:
V (gas) -?

1) 2Al + 2KOH + 6H2O => 2K [Al (OH) 4] + 3H2 ↑;
2) m solution (KOH) = ρ solution (KOH) * V solution (KOH) = 1.25 * 100 = 125 g;
3) m (KOH) = ω (KOH) * m solution (KOH) / 100% = 40% * 125/100% = 50 g;
4) M (KOH) = Mr (KOH) = Ar (K) * N (K) + Ar (O) * N (O) + Ar (H) * N (H) = 39 * 1 + 16 * 1 + 1 * 1 = 56 g / mol;
5) n (KOH) = m (KOH) / M (KOH) = 50/56 = 0.893 mol;
6) n (H2) = n (KOH) * 3/2 = 0.893 * 3/2 = 1.34 mol;
7) V (H2) = n (H2) * Vm = 1.34 * 22.4 = 30.02 l.

Answer: The volume of H2 is 30.02 liters.