$20 mL$ of ethane, $C X_{2} H X_{6}$ , is mixed with $100 mL$ of $O X_{2}$ and exploded. After cooling to room temperature the residual gas is passed through KOH solution. The volume of gas remaining after KOH treatment in mL (nearest integer, all volumes at the same room temperature and pressure) is

Numerical answer type, answer in mL, tolerance 0.5.

Answer: 30 mL

2 C X_{2} H X_{6} + 7 O X_{2} 4 CO X_{2} + 6 H X_{2} O (l)

20 mL

C X_{2} H X_{6}

needs

70 mL

O X_{2}

and gives

40 mL

CO X_{2}

. After cooling:

O X_{2}

left

= 100 - 70 = 30

CO X_{2}

= 40

. KOH removes the

CO X_{2}

, leaving

30 mL

O X_{2}