A liquid fuel. $C_{7} H_{16}$ is burnt with 10% deficit air than the stoichiometric air requirement. Assuming no hydrocarbons in the product, the volume % of $N_{2}$ in the product is
71.3

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Solution

The correct option is A 71.3
Perfect combustion:
$C_{7} H_{16} + 11 (O_{2} + \frac{79}{21} N_{2}) \to 8 H_{2} O + 7 C O_{2} + \frac{11 \times 79}{21} N_{2}$

10% deficit air:

$C_{7} H_{16} + 9.9 (O_{2} + \frac{79}{21} N_{2}) \to 8 H_{2} O + α C O_{2} + β C O + (\frac{9.9 \times 79}{21}) N_{2}$

Volume % of $N_{2}$ in the product $= \frac{(\frac{9.9 \times 79}{21})}{8 + 4.8 + 2.2 + (\frac{9.9 \times 79}{21})} = 0.713 = 71.3$

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