How many liters of hydrogen are released when 46 grams of ethyl alcohol interacts with potassium?

When potassium metal interacts with ethyl alcohol (ethanol), potassium ethylate is synthesized and hydrogen gas is released. The reaction is described by the following equation:

C2H5OH + K = C2H5OK + 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.

The amount of alcohol substance is.

M C2H5OH = 12 x 2 + 6 + 16 = 46 grams / mol;

N C2H5OH = 46/46 = 1 mol;

During this reaction, 1/2 = 0.5 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 = 0.5 x 22.4 = 11.2 liters;