How many litres of oxygen (at STP) is required for complete combustion of $39$ g of liquid benzene?

[Given that atomic weight of C $=$ 12, H $=$ 1, O $=$ 16]

84

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22.4

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42

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11.2

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Solution

The correct option is A $84$
The balanced chemical equation is as follows:

$C_{6} H_{6} (l) + 15 / 2 O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (g)$

$1$ mol ( $78$ g) of benzene reacts with $\frac{15}{2}$ moles $(\frac{15}{2} \times 22.4 L)$ of oxygen at STP for complete combustion.

Thus, complete combustion of $39$ g ( $0.5$ moles) of benzene will require $\frac{\frac{15}{2} \times 22.4 L}{78} \times 39 = 84 L$ of oxygen at STP.

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Similar questions

S T P

39 g m s

C = 12, H = 1, O = 16

(C_{6} H_{6})

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

39 g

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

39 g

C = 12, O = 16

C_{6} H_{6}

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