$500 m L$ of a gaseous hydrocarbon when burnt in excess of $O_{2}$ gave $2.5 L$ of $C O_{2}$ and $3.0 L$ of water vapour under same conditions. The molecular formula of the hydrocarbon is :

C_{4} H_{8}

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C_{4} H_{10}

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C_{5} H_{6}

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C_{5} H_{12}

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Solution

The correct option is C $C_{5} H_{12}$
The balanced chemical equation for the combustion reaction is
$C_{x} H_{y} 0.5 L 1 L + O_{2} \to x C O_{2} 2.5 L 5 L + \frac{y}{2} H_{2} O 3 L 6 L$
The volume is directly proportional to the number of moles.
Thus, 1 mole of hydrocarbon on combustion will give 5 moles of $C O_{2}$ and 6 moles of $H_{2} O$ .
Thus, the molecular formula of the hydrocarbon is $C_{5} H_{12}$ .

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500 mL of a gaseous hydrocarbon when burnt in excess of O2 gave 2.5 L of CO2 and 3.0 L of water vapour under same conditions. The molecular formula of the hydrocarbon is :

$500 m L$ of a gaseous hydrocarbon when burnt in excess of $O_{2}$ gave $2.5 L$ of $C O_{2}$ and $3.0 L$ of water vapour under same conditions. The molecular formula of the hydrocarbon is :