10 mL of a gaseous hydrocarbon CxHy is…

10 mL of a gaseous hydrocarbon CxHy is mixed with 60 mL of oxygen and the mixture is exploded in a closed eudiometer. After cooling to the initial…

Numerical answer

10\ \mathrm{mL}

of a gaseous hydrocarbon

\mathrm{C_xH_y}

is mixed with

60\ \mathrm{mL}

of oxygen and the mixture is exploded in a closed eudiometer. After cooling to the initial temperature (water condenses to liquid and its volume is negligible) the total volume is

45\ \mathrm{mL}

. On passing the residual gas through aqueous

\mathrm{KOH}

, the volume decreases to

25\ \mathrm{mL}

. All volumes are measured at the same temperature and pressure and the hydrocarbon burns completely. The value of

y

Show answer and solution

Answer: 6

Contraction by

\mathrm{KOH}

is the volume of

\mathrm{CO_2}

45-25=20\ \mathrm{mL}

. From

10\ \mathrm{mL}

of hydrocarbon this means

x=2

.
Residual gas after

\mathrm{KOH}

is excess

\mathrm{O_2}=25\ \mathrm{mL}

, so oxygen consumed

=60-25=35\ \mathrm{mL}

.
Combustion:

\mathrm{C_xH_y+(x+y/4)O_2\to xCO_2+\tfrac y2H_2O(l)}

. For

10\ \mathrm{mL}

10\left(x+\dfrac y4\right)=35\Rightarrow x+\dfrac y4=3.5

.
With

x=2

\dfrac y4=1.5

, so

y=6

(the hydrocarbon is ethane,

\mathrm{C_2H_6}

).
Check: total volume after cooling

=\mathrm{CO_2}+\mathrm{O_2}

left

=20+25=45\ \mathrm{mL}

, in agreement with the data given. Hence

y=6

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