Problem 2: 9.13 A stage extraction process is depicted in Fig. 19.13. In such systems, a stream containing a weight fraction y, of a chemical enters from the left at a mass flow rate of F. Simultaneously, a solvent carrying a weight fraction of the same chemical enters from the right at a flow rate of F. Thus, for stage i, a mass balance can be represented as FX+F341=FX+Fxx (P9.13a) At each stage, an equilibrium is assumed to be established between y, and x, as in (P9.136) where K is called a distribution coefficient. Equation (P9.136) can be solved for x and substituted into Eq. (P9.13a) to yield ,a --(1+x) +(6x) 3+1 = 0 (29.130) If Fs = 400 kg/h, y = 0.1, F3 = 800 kg/h, X.= 0, and K = 5, determine the values of you and Xif a five-stage reactor is used. Note that Eq. (P9.13.c) must be modified to account for the inflow weight fractions when applied to the first and last stages. 3 FIGURE.13 Atopom