20 ml of a hydrocarbon requires 100ml of oxygen under the same conditions for complete combustion 60 ml of $C O_{2}$ is formed. The formula of hydrocarbon is:

C_{2} H_{4}

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C_{3} H_{6}

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C_{3} H_{8}

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C_{2} H_{6}

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Solution

The correct option is C $C_{3} H_{8}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$
$20 m l$ $100 m l$ $60 m l$

$x = \frac{60}{20} = 3$

$x + \frac{y}{4} = \frac{100}{20} = 5$

$\Rightarrow x + \frac{y}{4} = 5$

$\Rightarrow 3 + \frac{y}{4} = 5$

$\Rightarrow \frac{12 + y}{4} = 5$

$\Rightarrow 12 + y = 20$

$y = 8$

Hence, the compound formed will be $C_{3} H_{8}$ .

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20 ml of a hydrocarbon requires 100ml of oxygen under the same conditions for complete combustion 60 ml of CO2 is formed. The formula of hydrocarbon is:

The correct option is C C3H8CxHy+(x+y4)O2→xCO2+y2H2O 20 ml 100 ml 60 mlx=6020=3 x+y4=10020=5 ⇒x+y4=5 ⇒3+y4=5 ⇒12+y4=5 ⇒12+y=20 y=8Hence, the compound formed will be C3H8.

20 ml of a hydrocarbon requires 100ml of oxygen under the same conditions for complete combustion 60 ml of $C O_{2}$ is formed. The formula of hydrocarbon is: