# How many liters of hydrogen react with a mixture of gases of 30% ethylene, 30% propene

**How many liters of hydrogen react with a mixture of gases of 30% ethylene, 30% propene and 40% butene, weighing 50 g?**

Let’s find the mass of gases in the mixture (ethylene, propene).

50 g – 100%,

m g – 30%.

m = (30% × 50 g): 100% = 15 g.

Let’s find the mass of butene.

50 g – 100%,

m g – 40%.

m = (40% × 50 g): 100% = 20 g.

Let’s find the amount of substance gases.

M (C2H4) = 28 g / mol.

n = 15 g: 28 g / mol = 0.536 mol.

M (C3H6) = 42 g / mol

n (C3H6) = 15 g: 42 g / mol = 0.357 mol.

M (C4H8) = 56 g / mol.

n = 20 g: 56 g / mol = 0.357 mol.

C2H4 + H2 = C2H6.

C3H6 + H2 = C3H8.

C4H8 + H2 = C4H10.

For 1 mole of each gas, there is 1 mole of H2.

The amount of substance gases and H2 will be equal.

Find the volume of hydrogen reacted with propene and ethylene.

V (H2) = 0.536 mol × 22.4 = 12 L.

V (H2) = 0.357 × 22.4 = 8 liters.

V (H2) = 0.357 × 22.4 = 8 liters.

Let’s find the total volume of hydrogen.

V (H2) = 12 l + 8 l + 8 l = 28 l.

Answer: 28 l.