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Standard XII

Chemistry

Stoichiometric Calculations

A mixture of ...

Question

A mixture of ethane and neon having a volume of $20$ ml, requires $49$ ml of oxygen for complete combustion. The percentage of ethane in the mixture is:

70 %

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30 %

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35 %

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64 %

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Solution

The correct option is A $70 %$
Neon is an inert gas and does not react with ethane or oxygen.
The reaction of combustion of ethane is $C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$ .
Thus, one mole ( $22.4$ L) of ethane requires $(\frac{7}{2} \times 22.4 = 78.4)$ L of oxygen for combustion.
Hence, $49$ ml of oxygen will combine with $49 \times \frac{22.4}{78.4} = 14$ ml of ethane.
Thus, the percentage of ethane in the mixture is $\frac{14}{20} \times 100 = 70 %$ .

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A mixture of ethane and neon having a volume of 20 ml, requires 49 ml of oxygen for complete combustion. The percentage of ethane in the mixture is:

A mixture of ethane and neon having a volume of $20$ ml, requires $49$ ml of oxygen for complete combustion. The percentage of ethane in the mixture is: