Calculate volume of cascade consisting of two ideal tank reactors required for the

production of $50M\frac{g}{?}$ day ethyl acetate. The reaction is described by the equation:

$C_2{H}_{5}OH+C{H}_3\frac{C}{O}OH=C{H}_3\frac{C}{O}O{C}_2{H}_5+H_{2}O$

$A+B=C+D$

Reaction rate in the liquid phase is defined by the expression:

$r=k(c_A{c}_B-\frac{c_C{c}_D}{K_C})$

Reaction rate constant equals $k=7\mathrm{.}93*{10}^{-6}[\frac{m^3}{\text{k}\text{m}\text{o}\text{l}*s}]$ and the equilibrium constant

equals $K_C=2\mathrm{.}93.$ The raw material is as follows $(\%$ mass$)$: $u_{A0}=46\%,u_{B0}=23\%,u_{D0}=$

$31\%.$ The reaction should be carried out to ${}_B=0\mathrm{.}35.$ Density of the mixture equals

$1020k\frac{g}{m^3}.$ Volumes of both reactors are the same.conversion equal to $0\mathrm{.}22.$Check also the volumetric flow rate value.Try to obtain a formula for space time and conversion $($which you use for each reactor$)$ from the balance of reactant B: FB$0-$FB$1+$rB$1*$V$=0$ for the first reactor and FB$1-$FB$2+$rB$2*$V$=0$ for the second reactor.megagrams not miligrams per day