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Q.
If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of $1 \,kg$ of it will be
AMUAMU 2002
Solution:
Combustion of butane takes place as
$C _{4} H _{10}+\frac{13}{2} O _{2} \longrightarrow 4 CO _{2}+5 H _{2} O $
$4 \times 12+10 \times 1 $
$=48+10$
$= 58\, g$
Now, $58 \,kg C _{4} H _{10}$ require $O _{2}$
$=\frac{13}{2} \times 32=208 \,kg$
$\therefore 1 \,kg\, C _{4} H _{10}$ require $O _{2}$
$=\frac{208}{58}=3.58 \,kg$