Calculate the volume of hydrogen released by the interaction of sodium with a mass of ethyl alcohol with a mass of 300 g.

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 = 300/46 = 6.522 mol;

During this reaction, 6.522 / 2 = 3.261 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 = 3.261 x 22.4 = 73.046 liters;