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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