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Mole Concept

Liquid benzen...

Question

Liquid benzene burns in oxygen according to $2 C_{6} H_{6 (l)} + 15 O_{2 (g)} \to 12 C O_{2 (g)} + 6 H_{2} O_{(l)}$ . How many litres of oxygen are required for complete combustion of $39 g$ of liquid benzene ?
(Atomic weight of $C = 12, O = 16$ )

11.2

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22.4

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42

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84

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Solution

The correct option is D $84$
The reaction is $2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$ .
The molecular weight of liquid benzene is $78$ g per mole. Thus, $39$ g of liquid benzene corresponds to $0.5$ moles.
$2$ moles of liquid benzene require $15$ moles of oxygen for combustion. Thus, $0.5$ moles of liquid benzene will require $0.5 \times \frac{15}{2} = 3.75$ moles of oxygen.
$1$ mole of oxygen at STP will occupy $22.4$ L. Hence, $3.75$ moles of oxygen will occupy $3.75 \times 22.4 = 84$ L at STP.

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Similar questions

(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

(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

(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)

O_{2}

(C_{6} H_{6})

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

O_{2}

39

S T P

39 g m s

C = 12, H = 1, O = 16

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