$15 mL$ of a gaseous hydrocarbon was exploded with $60 mL$ of $O X_{2}$ . The residual gas after cooling measured $45 mL$ , and after treatment with KOH it measured $15 mL$ . The hydrocarbon is

Answer: A

CO X_{2}

= 45 - 15 = 30 mL

, so

x = 30/15 = 2

O X_{2}

used

= 60 - 15 = 45 mL

15 (x + y /4) = 45

2 + y /4 = 3

y = 4

. The hydrocarbon is

C X_{2} H X_{4}