How many litres of oxygen measured at STP are required for complete combustion of 39 gms of liquid $C_{6} H_{6}$ ?

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Solution

molecular wt of $C_{6} H_{6} = 6 \times 12 + 1 \times 6 = 84$
$m o l e s = \frac{w t}{m o l w t} = \frac{39}{84} = 0.46$
$1 C_{6} H_{6} + 7.5 O_{2} \to 6 C O_{2} + 3 H_{2} O$
1 mole of $C_{6} H_{6}$ needs 7.5 moles of $O_{2}$
$\Rightarrow$ 0.46 mol of $C_{6} H_{6}$ requires $7.5 \times 0.46$ of $O_{2}$
$7.5 \times 0.46$ of $O_{2}$ occupies $7.5 \times 0.46 \times 22.4 l i t = 77.28 l i t$ .

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