Amount (in grams) of oxygen required for the complete combustion of 2 moles of benzene is.

450g

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480g

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500g

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530g

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Solution

The correct option is B 480g
The balanced reaction for combustion of benzene is :
$2 C_{6} H_{6} + 15 O_{2} \to$ $12 C O_{2} + 6 H_{2} O$
Therefore, 15 moles of oxygen gas is required for the complete combustion of benzene.
We know that, $1 m o l e$ of $O_{2}$ = $32 g$
$∴$ $15 m o l e s$ of $O_{2}$ = $15 \times 32$ = $480 g$

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Similar questions

How many grams of oxygen are required for complete combustion of 2 moles of benzene?

d m^{3}

(C_{6} H_{6})

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} (g)

117 g

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