Calculate the volume of hydrogen that will be released when sodium weighing 4.6 g interacts with ethyl alcohol.

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.

Let’s calculate the amount of sodium substance.

M Na = 23 grams / mol; N Na = 4.6 / 23 = 0.2 mol;

During this reaction, 0.2 / 2 = 0.1 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 = 0.1 x 22.4 = 2.24 liters;