Na weighing 12 g was placed in ethyl alcohol. How much hydrogen will be released in this case?

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 = 12/23 = 0.522 mol;

During this reaction, 0.522 / 2 = 0.261 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 = 0.261 x 22.4 = 5.846 liters;

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