What mass of acetic acid reacted with magnesium if 44.8 liters of hydrogen were released?

To solve the problem and make calculations, we write the equation:
Mg + 2CH3COOH = (CH3COO) 2Mg + H2 – substitution reaction, hydrogen is released;
M (CH3COOH) = 60 g / mol;
M (H2) = 2 g / mol.
Determine the number of moles of hydrogen:
1 mol of gas at n. y – 22.4 l;
X mol (H2) -44.8 L. hence, X mol (H2) = 1 * 44.8 / 22.4 = 2 mol.
Using the equation, we determine the amount of mol of acetic acid:
X mol (CH3COOH) – 2 mol (H2);
-2 mol – 1 mol from here, X mol (CH3COOH) = 2 * 2/1 = 4 mol.
We find the mass of acetic acid:
m (CH3COOH) = Y * M = 4 * 60 = 240 g.
Answer: the mass of the acid is 240 g.