How much ethane by weight is needed to obtain 5.6 liters of ethylene?

To solve, we write down the equation of the process:

C2H6 = H2 + C2H4 – dehydrogenation, ethylene is released;

We make calculations using the formulas:
M (C2H6) = 30 g / mol;

M (C2H4) = 28 g / mol.

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

X mol (C2H4) – 5.6 liters from here, X mol (C2H4) = 1 * 5.6 / 22.4 = 0.25 mol.

Y (C2H6) = 0.25 mol since the amount of substances is 1 mol.

4. Find the mass of the original substance:

m (C2H6) = Y * M = 0.25 * 30 = 7.5 g

Answer: you need ethane weighing 7.5 g