Feed stream flowing at $10mo\frac{l}{h}$ consists of $90\%Zr,$ and $10\%U$ is reacted with $HCl$ to produce metal chlorides. The $HCl$ stream contains some water in addition to $HCl$; the stream is a result of merging recycle stream and makes up pure $HCl$ stream via the following reactions:

$U+3HClUC{l}_3+\frac{3}{2}{H}_2$

$Zr+4HClZrC{l}_4+2H_2$

$U$ and $Zr$ are completely converted to chlorides. The total $HCl$ fed to the first reactor $($not the fresh feed$)$ is twice the amount required by the reaction stoichiometry. The $UC{l}_3$ produced is a solid and, hence, is readily removed from the remaining gaseous reaction products. The vapors from this first reactor are passed to a second reactor where the $ZrC{l}_4$ vapor is reacted with steam to produce solid $Zr{O}_2$ according to the following reaction:

$ZrC{l}_4+2H_{2}OZr{O}_2+4HCl$

The reaction goes to completion, and the $Zr{O}_2$ solid is separated from the gaseous reactor products. The gases leaving the second reactor are sent to an absorber$,$ which uses a solvent consisting of $90\%H_{2}O$ and $10\%HCl$ solution $($coming from stripper$)$ to absorb $HCl.$ The liquid leaving bottom of absorber containing $50\%HCl$ and $50\%H_{2}O$ and an overhead gas containing $90\%H_2$ and $10\%HCl.$ The absorber liquid is sent to a stripper, which boils off most of the $HCl$ to produce $90\%HCl$ and $10\%H_{2}O$ recycle vapor. The remaining stripper bottoms are cooled and recycled to the absorber for use as a solvent. Calculate the absorber inlet and exit solvent molar flow rates. $n_8=80\mathrm{.}92\frac{\text{m}\text{o}\text{l}}{h},n_9=161\mathrm{.}852\frac{\text{m}\text{o}\text{l}}{h}$