How much water is required to interact with 2.24 L of acetylene? What is the mass of the reaction product?

Based on the task data, we write down the process equation:
С2Н2 + Н2О = Н2С = СНОН (vinyl alcohol) = СН3 – СОН – hydration, acetaldehyde is released;

Calculations of the molar masses of substances:
M (C2H2) = 26 g / mol; M (H2O) = 18 g / mol; M (C2H4O) = 44 g / mol.

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

X mol (C2H2) -2.24 L from here, X mol (C2H2) = 1 * 2.24 / 22.4 = 0.1 mol;

Y (H2O) = 0.1 mol; Y (C2H4O) = 0.1 mol since the amount of substances is 1 mol.

We find the masses of water, aldehyde:
m (H2O) = Y * M = 0.1 * 18 = 1.8 g;

m (C2H4O) = Y * M = 0.1 * 44 = 4.4 g

Answer: you need water weighing 1.8 g, acetaldehyde weighing 4.4 g is obtained