What volume of hydrogen is required to hydrogenate 40 g of a mixture of propene and butene

What volume of hydrogen is required to hydrogenate 40 g of a mixture of propene and butene, the hydrogen density of which is 23.45?

Given:
mixture – C3H6, C4H8
m (mixture) = 40 g
D H2 (mixture) = 23.45

To find:
V (H2) -?

1) Write the reaction equations:
C3H6 + H2 => C3H8;
C4H8 + H2 => C4H10;
2) Calculate the molar mass of the mixture:
M (mixtures) = D H2 (mixtures) * M (H2) = 23.45 * 2 = 46.9 g / mol;
3) Calculate the amount of substance in the mixture:
n (mixture) = m (mixture) / M (mixture) = 40 / 46.9 = 0.853 mol;
4) Determine the amount of substance H2:
n (H2) = n (mixture) = 0.853 mol;
5) Calculate the volume H2:
V (H2) = n (H2) * Vm = 0.853 * 22.4 = 19.11 liters.

Answer: The volume of H2 is 19.11 liters.