How many liters of hydrogen will be obtained by the action of 2.3 g of ethyl alcohol with sodium maetall?

When sodium metal interacts with ethyl alcohol (ethanol), sodium ethylate is synthesized and hydrogen gas is released. The reaction is described by the following equation:

C2H5OH + Na = C2H5ONa + ½ H2;

1 mole of alcohol reacts with 1 mole of metal. This synthesizes 1 mol of ethylate and 0.5 mol of gaseous hydrogen.

The amount of alcohol substance is.

M C2H5OH = 12 x 2 + 6 + 16 = 46 grams / mol;

N C2H5OH = 2.3 / 46 = 0.05 mol;

In the course of this reaction, 0.05 / 2 = 0.025 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 = 0.025 x 22.4 = 0.56 liters;