$10$ ml of a gaseous hydrocarbon on combustion, gives $40$ ml of $C O_{2}$ gas and $50$ ml of $H_{2} O$ vapour at STP. The hydrocarbon is:

C_{4} H_{6}

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C_{8} H_{10}

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C_{4} H_{8}

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C_{4} H_{10}

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Solution

The correct option is A $C_{4} H_{10}$
The combustion reaction can be represented as $C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$ .

At STP, $1$ mole of any gas occupies $22.4 L$

$10$ ml of hydrocarbon, $50$ ml of water and $40$ ml of carbon dioxide corresponds to $0.45$ , $2.23$ and $1.79$ moles respectively.

Hence, the combustion of $1$ mole of hydrocarbon will produce $5$ moles of water and $4$ moles of carbon dioxide.

Thus, the molecular formula contains $4$ carbon atoms and $10$ hydrogen atoms.

The hydrocarbon is $C_{4} H_{10}$ and the combustion reaction is $C_{4} H_{10} + 6.5 O_{2} \to 4 C O_{2} + 5 H_{2} O$ .

So, the correct option is $D$

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10 ml of a gaseous hydrocarbon on combustion, gives 40 ml of CO2 gas and 50 ml of H2O vapour at STP. The hydrocarbon is:

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