Page 117 - Elements of Chemical Reaction Engineering 3rd Edition
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                               See. 3.3   Stoichiclmetric Table

                                     NaOH +  (C17 H,,COO),C,  H,   __j C,,H,,COONa  +  C, H,(OH),
                Choosing a basis
                   of calculation
                                       A   +        JB        *  C               +     fD
                                                    3
                                 We may then perfom the calculations shown in Table E3-4.1. Because this is a liq-
                                 uid-phase reaction, the density p is considered to be constant; therefore, V = V,>.










                                        TABLE E3-4.1.   STOICHIOMETRIC TABLE FOR LIQUID-PHASE SOAP REACTION
                                      Species    Symbol  Initially   Change   Remaining   Concentration

                                  NaOlH




                   Stoichiometric
                   . table (batch)








                                 Example 3-5  What Is the Limiting Reactant?

                                 Having set up thle stoichiometric table in Example 3-4, one can now readily use it to
                                 calcullate the  concentrations at  a  given  conversion. If  the  initial mixture consists
                                 solely of sodium hydroxide at a concentration of  10 mol/L (Le., 10 mol/dm3 or  10
                                 kmoil’m3 ) and of glyceryl stearate at a concentration of 2 g mol/L, what is the con-
                                 centration of  glycerine when  the conversion of  sodium hydroxide is (a) 20%  ,and
                                  (b) 90%?

                                  Solution
                                  Only the reactants NaOH  and  (CI7 H3,COO),C,  H,  are initially present; therefore,
                                  0, = 0, = 0.
                                  (a) For 20% conlversion:


                                        CD = CAo(:]   = (10)k) 0.67 g moVL = 0.67 mol/dm3
                                                             =

                                                                     = lO(0.133) = 1.33moVdm3
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