Toluene, $C_{7} H_{8}$ , is added to gasoline to increase its octane rating. What is the volume ratio of air to toluene vapor to burn completely to form ${CO}_{2}$ and $H_{2} O$ ? (Assume air is $20 % O_{2}$ by volume.)

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$\frac{28}{1}$

$\frac{11}{1}$

$\frac{45}{1}$

$\frac{9}{1}$

answer is D.

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Detailed Solution

In the following reaction,

$C_{7} H_{8} + 9 O_{2} ⟶ 7 {CO}_{2} + 4 H_{2} O$

1 mole toluene need 9 moles $O_{2}$ 9n one mole air, only $20 % (0.2$ mole) is $O_{2}$ .

It will be present in 45 moles of air,

$\frac{a i r}{c_{7} H_{8}} = \frac{45}{1}$

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