How many litres of oxygen at $S T P$ are required for complete combustion of $39 g m s$ of liquid Benzene?

[Atomic weights $C = 12, H = 1, O = 16$ ]

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 C $84$
Molar mass of $C_{6} H_{6} = 78 g m$
$39 g m s$ of $C_{6} H_{6} (l) = \frac{39}{78}$ moles $=$ 0.5 moles
$2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (l)$
For 2 moles, we require 15 moles of $O_{2}$ , for 0.5 mole we will require,
$=> \frac{0.5}{2} \times 15 = \frac{15}{4}$ moles of $O_{2}$
$=> \frac{15}{4} \times 22.4$ litres $=$ 84 litres

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Similar questions

39

=

=

=

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})

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}

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