You visited us 1 times! Enjoying our articles? Unlock Full Access!

Real Gases

Liquid benzen...

Question

Liquid benzene $(C_{6} H_{6})$ burns in oxygen according to the following reaction:
$2 C_{6} H_{6} (l) + 15 O_{2} (g) ⟶ 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of $O_{2}$ at STP are needed for complete combination of $39$ grams of liquid benzene?

11.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

74

No worries! We‘ve got your back. Try BYJU‘S free classes today!

84

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

22.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $84$
The balanced reaction is as follows:
$2 C_{6} H_{6} + 15 O_{2} ⟶ 12 C O_{2} + 6 H_{2} O$
$2 \times 78$ g of $C_{6} H_{6}$ $= 15 \times 22.4$ litres of $O_{2}$ at STP
$39$ g of $C_{6} H_{6} = \frac{15 \times 22.4 \times 39}{2 \times 78} = 84$ litres of $O_{2}$ at STP
Hence, $84$ L of $O_{2}$ is needed for complete combustion of $39$ g of liquid benzene.

Suggest Corrections

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)

O_{2}

(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} 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

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

39

C_{6} H_{6}

Join BYJU'S Learning Program