The composition of LPG is butane and isobutane. The amount of oxygen that would be required for combustion of $1 k g$ of LPG will be approximately

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$4.5 K g$

$2.7 K g$

$1.8 K g$

$3.6 K g$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The combustion reaction of butane is as follows:

$\underset{1 m o l}{C_{4} H_{10} (g)} + \frac{13}{2} O_{2} (g) \to 4 {C O}_{2} (g) + 5 H_{2} O (l)$

From the above reaction we can say that (58g) of $C_{4} H_{10}$ will require $(\frac{13}{2} \times 32 = 208 g)$ of $O_{2}$

$∴ O_{2}$ required for $1 k g$ butane

$= \frac{208}{58} \times 1000 g$

= $3.58 \times 10^{3} g$

Hence the correct option is (D).

JEE

NEET

Foundation JEE

Foundation NEET

CBSE