The formula of a gaseous hydrocarbon which requires 6 times of its own volume of $O_{2}$ for complete oxidation and produces 4 times its own volume of $C O_{2} is C_{x} H_{y}$ . The value of $y$ is (nearest integer):

8.00

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8.0

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Solution

$C_{x} H_{y} (g) + (x + \frac{y}{4}) O_{2} (g) \to x C O_{2} (g) + \frac{y}{2} H_{2} O (l)$
Let $a$ litres of $C_{x} H_{y}$ be used
Given:
$a \times (x + \frac{y}{4}) = 6 a . . . (1)$
and
$a x = 4 a . . . (2)$
So,
$4 + \frac{y}{4} = 6$
$\Rightarrow y = 8$
From equation $(1)$ , $x = 4$
Hence hydrocarbon is $C_{4} H_{8}$

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The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is (nearest integer):

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The formula of a gaseous hydrocarbon which requires 6 times of its own volume of $O_{2}$ for complete oxidation and produces 4 times its own volume of $C O_{2} is C_{x} H_{y}$ . The value of $y$ is (nearest integer):