Question

LPG is a mixture of n-butane & iso-butane. What is the volume of Oxygen required to burn $1\mathrm{kg}$ of LPG at NTP?

• A. $500\mathrm{L}$
• B. $1000\mathrm{L}$
• C. $2240\mathrm{L}$
• D. $2690\mathrm{L}$

Answer 1

The correct option is D

$2690\mathrm{L}$

The correct option is (D):

Explanation for the correct answer:

Step 1: Finding the molecular weight of the compounds

${\mathrm{C}}_4{\mathrm{H}}_{10}+\frac{13}{2}{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+5{\mathrm{H}}_2\mathrm{O}$

The molecular weight of ${\mathrm{C}}_4{\mathrm{H}}_{10}$ = $4\left(12\right)+10\left(1\right)=58\mathrm{gm}$

The molecular weight of ${\mathrm{O}}_2$ = $2\left(16\right)=32\mathrm{gm}$

Step 2: Finding the volume of Oxygen required.

$58\mathrm{gm}{\mathrm{CH}}_4$ required volume of ${\mathrm{O}}_2$ at NTP

= $\frac{13}{2}\times 24\mathrm{lit}$

Then , $1\mathrm{kg}\mathrm{of}{\mathrm{CH}}_4$ requires

= $\frac{13\times 1000}{2\times 58}\times 24\mathrm{lit}$

= $2690\mathrm{L}$

So, The correct answer is (D):- $2690\mathrm{L}$