Question

1. LPG is a mixture of n-butane and iso-butane. The volume of oxygen needed to burn 1 kg of LPG at STP would be 1) 2240L 2) 2544L 3) 1000L 4) 500L​

Answer 1

Explanation:

Correct option is

B

2690 L

reaction:

C

4

H

10

+

2

13

O

2

→4CO

2

+5H

2

O

58gm

2

13

×32

58g CH

4

required volume of O

2

at NTP=

2

13

×24 lit

so, 1kg of CH

4

requires =

2×58

13×1000

×24 lit=2690L

Answer 2

Answer:

The volume of oxygen needed to burn 1 kg of LPG at STP would be 2690L i.e.none of the options are correct.

Explanation:

The reaction of burning between butane (both n-butane and iso-butane) is given as,

$C_4H_1_0+\frac{13}{2} O_2 \rightarrow 4CO_2+5H_2O$       (1)

C₄H₁₀=Butane

O₂=Oxygen

CO₂=Carbon dioxide

H₂O=water

The molecular weight of butane is =58g

From reaction (1) we can see that 58-gram C₄H₁₀ requires $\frac{13}{2} \times 22.4 L$ of O₂.

So, for 1kg of LPG, the required oxygen is,

$\frac{13\times 1000}{2\times 58} \times 24=2690L$

Hence, the volume of oxygen needed to burn 1 kg of LPG at STP would be 2690L i.e.none of the options are correct.