How many litres of oxygen measured at STP are required for complete combustion of 39 g of liquid $C_{6} H_{6}$ ?

11.2

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22.4

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Solution

The correct option is D 84
$C_{6} H_{6} + 7.5 O_{2} \to 6 C O_{2} + 3 H_{2} O$

The molar mass of $C_{6} H_{6}$ is 78 g/mol.

39 g of $C_{6} H_{6} = \frac{39 g}{78 g / m o l} = 0.5 m o l$

0.5 moles of $C_{6} H_{6}$ will react with $0.5 \times 7.5 = 3.75$ moles of $O_{2}$

At STP, 1 mole $O_{2} =$ 22.4 L

3.75 L $O_{2} = 3.75 \times 22.4 = 84$ L.

Hence, option D is correct.

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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

2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O

39

C_{6} H_{6}

(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}

S T P

39 g m s

C = 12, H = 1, O = 16

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