Volume of oxygen gas required at NTP to completely oxidise 5.8 g of butane $(C_{4} H_{10})$ is:

14.56 L

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11.2 L

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22.4 L

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10.25 L

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Solution

The correct option is A 14.56 L
$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O$

So , 1 Mole of $C_{4} H_{10}$ requires 6.5 moles of $O_{2}$
0.1 Mole of $C_{4} H_{10}$ requires 0.65 moles $O_{2}$

So , Volume of oxygen gas required is $0.65 \times 22.4$ = 14.56 L

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