How many liters of hydrogen will be released when an excess of sodium reacts with 8 g of methanol?

2CH3OH + 2Na = 2CH3ONa + H2 – substitutions, hydrogen is evolved;
Let’s make calculations using the formulas:
M (CH3OH) = 32 g / mol;

M (H2) = 2 g / mol;

Y (CH3OH) = m / M = 8/32 = 0.25 mol.

Proportion:
0.25 mol (CH3OH) – X mol (H2);

-2 mol -1 mol hence, X mol (H2) = 0.25 * 1/2 = 0.125 mol.

We find the volume of H2:
V (H2) = 0.125 * 22.4 = 2.8 L

Answer: 2.8 liters of hydrogen was obtained.