A CYLINDER CONTAINS EITHER ETHYLENE OR PROPYLENE 12 ML OF GAS REQUIRED 54 ML OF OXYGEN FOR COMPLETE COMBUSTION. THE GAS IS 
A) ETHYLENE 
B) PROPYLENE 
C) 1:1 MIXTURE OF TWO GASES

Dear Student,

First let's write the balanced equation for the complete combustion for ethylene and propylene gas to have a justified answer,

$$\begin{gathered} Ethylene reaction scheme is, \\ {C}_2{H}_4+ 2O_2\to 2H_{2}O+2C{O}_2 \\ Propylene reaction scheme is, \\ {C}_3{H}_6+4.5O_2\to 3H_{2}O+3C{O}_2 \end{gathered}$$

By using gas law at STP condition,lets compute for the volume of oxygen as below,

1 mole of ethylene reacts with 2 moles of oxygen
22400 ml = 2 (22400)ml

for 12 ml =?

12 ml(2 mole) = 24 ml

in similar way propylene reacts with 4.5 moles of oxygen the computation is shown below,

22400 ml = 4.5(22400) ml

for 12 ml=?

12ml(4.5 mole) = 54 ml

Hence the answer is Option (B) Propylene

Regards.