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Q.

40 ml of a hydrocarbon undergoes combustion in 260 ml of oxygen and gives 160 ml of $C O_{2}$ . If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is

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a

$C_{3} H_{8}$

b

$C_{4} H_{8}$

c

$C_{6} H_{14}$

d

$C_{4} H_{10}$

answer is D.

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Detailed Solution

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$
$40 C_{x} H_{y} 40 (x + \frac{y}{4}) O_{2} 40 x C O_{2} 40 \frac{y}{2} x_{2} O$
Given 40 $x$ = 160
$∴ x = 4$
$40 (4 + \frac{y}{4}) = 260$
y = 10
$∴$ formula of Hydrocarbon $= C_{4} H_{10}$

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