The volume of hydrogen that can be attached to 10 liters of propene?

Let’s execute the solution:

In accordance with the condition of the problem, we will write the data:
V = 10 l Chl -?

Н3С – СН = СН2 + Н2 = Н3С – СН2 – СН3 – addition, propane is released;

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

X mol (C3H6) -10 L from here, X mol (C3H6) = 1 * 10 / 22.4 = 0.45 mol;

Y (H2) = 0.45 mol since the amount of substances is 1 mol.

We find the volume of H2:
V (H2) = 0.45 * 22.4 = 10.08 L

Answer: you need hydrogen with a volume of 10.08 liters