Calculate the volume of hydrogen obtained by the interaction of 1.5 mol of methanol with metallic sodium

Calculate the volume of hydrogen obtained by the interaction of 1.5 mol of methanol with metallic sodium, taken in sufficient quantity, if the volume fraction of the product yield is 85% of the theoretically possible.

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.

N CH3OH = 1.5 mol;

Taking into account the reaction yield of 85% of the theoretical, 1.5 / 2 x 0.85 = 0.638 mol of hydrogen will be released during this reaction.

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 = 0.638 x 22.4 = 14.29 liters;