How many litres of $O_{2}$ at STP are needed for the complete combustion of $80 g$ of liquid hexane ( $C_{6} H_{14})$ ?
( $G i v e n : C_{6} H_{14} + O_{2} \to C O_{2} + H_{2} O$ )

55.6 L

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197.9 L

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94.6 L

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122.6 L

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Solution

The correct option is B $197.9 L$
$2 C_{6} H_{14} + 19 O_{2} \to 12 C O_{2} + 14 H_{2} O$

2 moles of hexane reacts with 19 moles of $O_{2}$

Molar mass of hexane = $12 \times 6 + 14 g = 86 g$

$86 \times 2$ g of hexane completely combusts in $19 \times 22.4$ L of oxygen at STP.

$172$ g of hexane completely combusts in $425.6$ L of oxygen at STP.

So, the amount of $O_{2}$ required to completely combust $80 g$ of hexane $= \frac{425.6}{172} \times 80 L of oxygen at STP$ .

= $197.9$ litres of $O_{2}$

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