Test Index

Hydrocarbons Part 1

Section: Chemistry

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Question : 79 of 100

Marks: +1, -0

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20%

\mathrm{O}_2

by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

$\mathrm{C}_3\mathrm{H}_8$
$\mathrm{C}_4\mathrm{H}_8$
$\mathrm{C}_4\mathrm{H}_{10}$
None of the above

Solution:

Volume of

\mathrm{O}_2 = 375 \times \frac{20}{100} = 75

ml

Volume of gaseous hydrocarbon = 15 ml

\mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O}(\mathrm{l})

15\left(x + \frac{y}{4}\right) = 75

....(1)

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

Excess air

= 375 - 75 = 300

300 + volume of

\mathrm{CO}_2 = 330

Volume of

\mathrm{CO}_2 = 30

15x = 30

x = 2

(Correct option is not given)

2 + \frac{y}{4} = 5

\frac{y}{4} = 3

y = 12

\mathrm{C}_2\mathrm{H}_{12}

(not possible)

Hence No answer is correct.

However, considering the equation no. 1 alone, if we put x and y values, 3 and 8 respectively, then equation no. 1 is satisfied and answer will be $\mathrm{C}_3\mathrm{H}_8$

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