A gaseous mixture of propane and butane of volume 3 L on complete combustion produces 10 L of $C{O}_2$ under standard conditions of temperature and pressure. The ratio of volumes of propane to butane is:

The correct option is A 2 : 1

Propane-

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

3 liters of propane would produce 9 liters of $C{O}_2$ ($3\times 3=9$)

Butane-

$C_4{H}_{1}0+65O_2\to 4C{O}_2+5H_{2}O$

3 liters of butane would produce 12 liters of $C{O}_2$ ($3\times 4 = 12$)

Liters of propane = X

Liters of butane = Y

$X+Y=3$

$3X+4Y=10$

Solve 2 simultaneous equations to get-

Y = 1

2 mole propane and 1 mole butane.So the ratio is 2:1.

Hence option B is correct.