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Stoichiometry

A mixture of ...

Question

A mixture of $H_{2}, N_{2}$ & $O_{2}$ occupying 100 ml, underwent reaction so as to form a $H_{2} O_{2}$ (l) and $N_{2} H_{2}$ as the only products, causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10 ml. To the remaining mixture, excess $H_{2}$ was added and the above reaction was repeated, causing a reduction in volume of 10 ml. Identify the composition of the initial mixture in ml (No other products are formed).

A

N_{2} = 30 m l, H_{2} = 40 m l

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B

N_{2} = 32 m l, H_{2} = 30 m l

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C

N_{2} = 36 m l, H_{2} = 46 m l

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D

N_{2} = 29 m l, H_{2} = 31 m l

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Solution

The correct option is A $N_{2} = 30 m l, H_{2} = 40 m l$
The volume percent of the hydrogen , nitrogen and oxygen in the mixture is 40 ml, 30 ml and 30 ml respectively.
This can be verified by the following reactions.
$2 H_{2} 40 m l + N_{2} 30 m l + O_{2} 30 m l \to + H_{2} O_{2} (l) 20 m l + N_{2} H_{2} 20 m l$
In this reaction, hydrogen is the limiting reagent. The volume of unreacted nitrogen and oxygen remaining is 10 ml each.
The volume contraction of 60 ml is because, the original 100 ml of the mixture now contains 10 ml nitrogen, 10 ml hydrogen and 20 ml $N_{2} H_{2}$ . $H_{2} O_{2}$ is a liquid now.
10 ml of oxygen dissolves in alkaline pyrogallol solution which gives a volume contraction of 10 ml.
The final volume contraction of 10 ml is due to the following reaction.
$H_{2} 10 m l + N_{2} 10 m l \to N_{2} H_{2} 10 m l$

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