When potassium metal interacts with methyl alcohol (methanol), potassium methoxide is synthesized and hydrogen gas is released. The reaction is described by the following equation:
CH3OH + K = CH3OK + 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.
The amount of alcohol substance is.
M CH3OH = 12 + 4 + 16 = 32 grams / mol;
N CH3OH = 50/32 = 1.563 mol;
In the course of this reaction, 1.563 / 2 = 0.782 mol of hydrogen will be released and 1.563 mol of potassium methoxide will be synthesized.
Let’s calculate its weight.
M CH3OK = 12 + 3 + 16 + 39 = 70 grams / mol;
m CH3OK = 1.563 x 70 = 109.41 grams;
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