Calculate percentage volume of butene in original gas mixture. | TestHub

Passage / Comprehension

A mixture of $H_{2}$ and butene $(C_{4} H_{8})$ was collected in a Eudiometer tube. Then, 80 ml of oxygen were also introduced. The resulting mixture of all the gases was exploded. After cooling resulting gaseous mixture passes through KOH solution which caused a contraction of 32 ml and 16 ml of oxygen alone were left behind.

Calculate percentage volume of butene in original gas mixture.

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Answer:

Solution:

KOH absorbs CO₂, so the contraction is equal to the CO₂ formed, which is $32 m L$ .

For the combustion of butene:

$C_{4} H_{8} + 6 O_{2} \to 4 C O_{2} + 4 H_{2} O$

One volume of butene gives 4 volumes of CO₂, so the butene volume is $\frac{32}{4} = 8 m L$ .

From previous calculations, the initial mixture volume was $40 m L$ .

The percentage of butene is calculated as:

$% b u t e n e = \frac{8}{40} \times 100 = 20%$

Correct option: (B) 20%

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