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Q.

500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of ${CO}_{2}$ and 3 litres of water vapours. All volume being measured at the same temperature and pressure. The formula of the hydrocarbon is :

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a

$C_{4} H_{10}$

b

$C_{4} H_{8}$

c

$C_{5} H_{10}$

d

$C_{5} H_{12}$

answer is B.

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Detailed Solution

The reaction takes place as

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x {CO}_{2} + \frac{y}{2} H_{2} O 500 \begin{array}{lll} - 500 x 500 \times \frac{y}{2} \\ - = 2500 = 3000 \\ \Rightarrow x = 5 y = 12 \end{array}$

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500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO2 and 3 litres of water vapours. All volume being measured at the same temperature and pressure. The formula of the hydrocarbon is :