When sodium metal interacts with methyl alcohol (methanol), sodium methoxide is synthesized and hydrogen gas is released. The reaction is described by the following equation:
CH3OH + Na = CH3ONa + ½ H2;
1 mole of alcohol reacts with 1 mole of metal. This synthesizes 1 mol of methylate and 0.5 mol of hydrogen gas.
Let’s calculate the amount of sodium substance.
M Na = 23 grams / mol; N Na = 7.5 / 23 = 0.326 mol;
Let’s calculate the amount of alcohol substance.
M CH3OH = 12 + 4 + 16 = 32 grams / mol; N CH3OH = 80/32 = 2.5 mol;
During this reaction, 0.326 / 2 = 0.163 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 = 0.163 x 22.4 = 3.651 liters;
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