$10 mL$ of a gaseous hydrocarbon $C X_{x} H X_{y}$ was exploded with $50 mL$ of $O X_{2}$ (in excess). After cooling to room temperature the gas volume was $35 mL$ , and after absorption in KOH it was $5 mL$ . The value of $y$ (an integer) is

Numerical answer type, tolerance 0.1.

Answer: 6

CO X_{2}

formed

= 35 - 5 = 30 mL

, so

x = 30/10 = 3

O X_{2}

left

= 5 mL

, so

O X_{2}

used

= 45 mL

. From

C X_{x} H X_{y} + (x + \frac{y}{4}) O X_{2} x CO X_{2} + (\frac{y}{2}) H X_{2} O (l)

10 (x + y /4) = 45

3 + y /4 = 4.5

y = 6

. The hydrocarbon is

C X_{3} H X_{6}