How much hydrogen will be released when sodium interacts with 100 g of methyl alcohol

How much hydrogen will be released when sodium interacts with 100 g of methyl alcohol containing 20% oil impurities?

Given:
m tech. (CH3OH) = 100 g
ω approx. = 20%

To find:
V (H2) -?

Decision:
1) 2CH3OH + 2Na => 2CH3ONa + H2 ↑;
2) M (CH3OH) = Mr (CH3OH) = Ar (C) + Ar (H) * 4 + Ar (O) = 12 + 1 * 4 + 16 = 32 g / mol;
3) ω (CH3OH) = 100% – ω approx. = 100% – 20% = 80%;
4) m clean. (CH3OH) = ω (CH3OH) * m tech. (CH3OH) / 100% = 80% * 100/100% = 80 g;
5) n (CH3OH) = m pure. (CH3OH) / M (CH3OH) = 80/32 = 2.5 mol;
6) n (H2) = n (CH3OH) / 2 = 2.5 / 2 = 1.25 mol;
7) V (H2) = n (H2) * Vm = 1.25 * 22.4 = 28 HP

Answer: The volume of H2 is 28 liters.