WHY THESE QUESTIONS ARE DIFFERENT IN SOLUTIONS BECAUSE EVERY TIME I TRY TO SOLVE IT IN ONE METHOD THE ANSWERS DIFFERS.Example $1$: The mixture that contains $30$ mole $\%n-$pentane and the remainder $n-$hexane is to be separated into $95\%$ pure pentane and $95\%$ pure hexane using a distillation column. The feed enters the column at a rate of $1000$Kmo$\frac{l}{h}.$

Calculate the outlets mole flow rate.

Solution

$=>F=D+W=>100=D+Wq,$ Equ$(1)$

:$.D=277\mathrm{.}7$kmo$\frac{l}{h}$

$=>,x_{F}F=x_{D}D+x_{W}W$

$0\mathrm{.}3(100)=0\mathrm{.}95D+0\mathrm{.}05Wq,$

$q,$ Equ$(2)$

:$.W=722\mathrm{.}22$kmo$\frac{l}{h}$

Example $2$: The mixture that contains $30$ mole $\%n-$pentane and the remainder $n-$hexane is fed to a distillation column to recover $90\%$ of the $n-$pentane in the top and $90\%n-$hexane in the bottom. The feed enters the column at a rate of $1000$Kmo$\frac{l}{h}.$ Calculate the outlets mole flow rate

Solution

$n-$pentane $(A)$ enter in feed $=0\mathrm{.}31000=300$kmo$\frac{l}{h}r$

$n-$hexane $(B)$ enter in feed $=0\mathrm{.}71000=700$kmo$\frac{l}{h}r$

$n-$pentane $(A)$ in distillate $=3000\mathrm{.}9=270$kmo$\frac{l}{h}r$

$n-$pentane $(A)$ in bottom $=3000\mathrm{.}1=30$kmo$\frac{l}{h}r$

$n-$hexane $(B)$ in bottom $=7000\mathrm{.}9=630$kmo$\frac{l}{h}r$

$n-$hexane $($B$)$ in distillate $=7000\mathrm{.}1=70$kmo$\frac{l}{h}r$