$20 mL$ of a gaseous hydrocarbon $C_{x} H_{y}$ is burnt completely with $100 mL$ of $O X_{2}$ (excess). After cooling to the initial temperature (water condenses to liquid) the gas volume is $70 mL$ , and after passing through aqueous KOH it falls to $30 mL$ . Find $y$ (integer).

Numerical answer type, tolerance 0.5.

Answer: 6

KOH absorbs

CO X_{2}

70 - 30 = 40

mL, so

x = 40/20 = 2

. Unused

O X_{2} = 30

mL, so

O X_{2}

used

= 70

mL. Per mL of hydrocarbon,

x + y /4 = 70/20 = 3.5

, so

y /4 = 1.5

and

y = 6