Question

LPG is a mixture of n - butane & iso - butane. The volume of oxygen needed to burn $464$ kg of LPG at NTP would be:

• A. $22400L$
• B. $2329.6L$
• C. $1164.8L$
• D. $582.4L$

Answer 1

The correct option is C $1164.8L$

$C_4{H}_{10\left(g\right)}+\frac{13}{2}{O}_{2\left(g\right)}\to 4C{O}_{2\left(g\right)}+5H_2{O}_{\left(g\right)}$
Given : weight of LPG = 464 g
Moles of LPG $=\frac{464}{58}=8$ mol
$1$ mole of LPG required oxygen $=\frac{13}{2}$ moles
$8$ mole of LPG required oxygen
$=\frac{13}{2}\times 8$ mol
= $52$ mole
Volume of oxygen required
$=\text{No of moles}\times 22.4L$
$=52\times 22.4=1164.8L$