Indicate the amount of hydrogen that will be released when 19.2 g of methanol interacts with an excess of sodium.

When sodium metal interacts with methyl alcohol (methanol), sodium methoxide is synthesized and hydrogen gas is released. The reaction is described by the following equation:
CH3OH + Na = CH3ONa + ½ H2;
1 mole of alcohol reacts with 1 mole of metal. This synthesizes 1 mol of methylate and 0.5 mol of hydrogen gas.
The amount of alcohol substance is.
M CH3OH = 12 + 4 + 16 = 32 grams / mol;
N CH3OH = 19.2 / 32 = 0.6 mol;
In the course of this reaction, 0.6 / 2 = 0.3 mol of hydrogen will be released.
Let’s calculate its volume. To do this, multiply the amount of substance by the volume of 1 mole of gas (22.4 liters).
V H2 = 0.3 x 22.4 = 6.72 liters;