We know, conbustion of Hydrocarbon

C_x H_y + (x + $${y \over 4}$$)O₂  $$ \to $$  x CO₂ + $${y \over 2}$$ H₂O

So, to consume 1 mole of C_x H_y we need

(x + $${y \over 4}$$) mole of O₂ gas.

As both C_x H_y and O₂ are gas,

So avogadro's law is applicable on them.

At constant temperature and pressure according to avogadro's law, volume $$ \propto $$ mole

So, for 5L of C_x H_y the

volume of O₂ = 5(x + $${y \over 4}$$)

According to the question,

5(x + $${y \over 4}$$) = 25

$$\therefore\,\,\,$$ (x + $${y \over 4}$$) = 5 . . . . . (1)

For Ethane (C₂ H₆), x = 2 and y = 6

$$\therefore\,\,\,$$ x + $${y \over 4}$$ = 2 + $${6 \over 4}$$ $$ \ne $$ 5

$$\therefore\,\,\,$$ C₂ H₆ can't be the required alkane.

For Propane (C₃ H₈), x = 3. and y = 8

$$\therefore\,\,\,$$ x + $${y \over 4}$$ = 3 + $${8 \over 4}$$ = 5

$$\therefore\,\,\,$$ Propane (C₃ H₈) is the right alkane.

Similarly for Butane (C₄ H₁₀) and Isobutane (C₄ H₁₀) you can check
x + $${y \over 4}$$ $$ \ne $$ 5.