How many grams of oxigen is essentially required for complete combustion of $3$ moles of butane gas ?

624

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312

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128

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64

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Solution

The correct option is D $312$ g
The combustion of butane is given by,
$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O$

Hence, For combustion of 1 mole of butane 6.5 moles of oxygen is required.
Therefore, For 3 moles of butane 19.5 moles of oxygen will be required,
Mass of oxygen required $= 19.5 m o l \times 16 g / m o l = 312 g$

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