JEE Advanced 2022 Paper 1 · Chemistry · Mole Concept and Stoichiometry · medium

A 30 mL gaseous mixture of methane, ethane and nitrogen is mixed with excess oxygen and sparked. After the products are cooled back to the initial…

Numerical answer

30\ \mathrm{mL}

gaseous mixture of methane, ethane and nitrogen is mixed with excess oxygen and sparked. After the products are cooled back to the initial temperature and pressure (water condenses completely), the volume is found to have decreased by

44\ \mathrm{mL}

. Treatment of the residual gas with concentrated KOH solution decreases the volume by a further

28\ \mathrm{mL}

. The volume, in

\mathrm{mL}

, of nitrogen in the original mixture is

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Answer: 10

Let the volumes of

\mathrm{CH_4}

and

\mathrm{C_2H_6}

x

and

y

mL. Combustion with liquid water formed:

\mathrm{CH_4+2O_2\to CO_2+2H_2O(l)}

gives a contraction of

1+2-1=2

mL per mL of methane;

\mathrm{C_2H_6+\tfrac72 O_2\to 2CO_2+3H_2O(l)}

gives

1+3.5-2=2.5

mL per mL of ethane. Nitrogen is inert and does not contribute. Hence

2x+2.5y=44

. The KOH absorbs only

\mathrm{CO_2}

x+2y=28

. Solving,

2x+4y=56

and subtracting gives

1.5y=12

, so

y=8

and

x=12

. Nitrogen

=30-12-8=10

mL. Check: contraction

=24+20=44

\mathrm{CO_2}=12+16=28

. Answer: 10.

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