The formula of a gaseous hydrocarbon which requires 6 times of its own volume of

Question:

The formula of a gaseous hydrocarbon which requires 6 times of its own volume of $\mathrm{O}_{2}$ for complete oxidation and produces 4 times its own volume of $\mathrm{CO}_{2}$ is $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}$. The value of $\mathrm{y}$ is

Solution:

(8)

$\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+6 \mathrm{O}_{2} \longrightarrow 4 \mathrm{CO}_{2}+\frac{\mathrm{y}}{2} \mathrm{H}_{2} \mathrm{O}$

Applying POAC on 'O' atoms $6 \times 2=4 \times 2+y / 2 \times 1$

$y / 2=4 \Rightarrow y=8$

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