How many grams of ethyl alcohol, reacting with a slight excess of potassium, release 2.5 liters of hydrogen.

Let’s implement the solution:

Let’s write the equation:

2C2H5OH + 2K = 2C2H5OK + H2 – substitutions, hydrogen is evolved;

Calculations by the formulas of substances:
M (C2H5OH) = 46 g / mol;

M (H2) = 2 g / mol.

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (H2) – – 2.5 liters from here, X mol (H2) = 1 * 2.5 / 22.4 = 0.1 mol;

X mol (C2H5OH) – 0.1 mol (H2);

-2 mol -1 mol hence, X mol (C2H5OH) = 2 * 0.1 / 1 = 0.2 mol.

We find the mass of alcohol:
m (C2H5OH) = Y * M = 0.2 * 46 = 9.2 g

Answer: the mass of ethyl alcohol is 9.2 g