Consider the following hypothetical reaction. Show yourwork to answer the follow

Question

Consider the following hypothetical reaction. Show yourwork to answer the following questions:                       A     +     B            ------->   C        MW 50g/mol         80g/mol                    98 g/mol a) If 0.4 g of A reacts with 0.6 g of B what isthe limiting reagent? b) What is the theoretical yield of C? c) What is the percent yield of the reaction if0.6 g of C was isolated upon completion of the reaction? Please SHOW yourwork. Consider the following hypothetical reaction. Show yourwork to answer the following questions:                       A     +     B            ------->   C        MW 50g/mol         80g/mol                    98 g/mol a) If 0.4 g of A reacts with 0.6 g of B what isthe limiting reagent? b) What is the theoretical yield of C? c) What is the percent yield of the reaction if0.6 g of C was isolated upon completion of the reaction? Please SHOW yourwork.

Explanation / Answer

           The hypothetical reaction is as follows:                 A      +              B            ------->   C          MW 50 g/mol        80g/mol                    98 g/mol     a) mass of A = 0.4 g              so moles presentin A = mass / MW                                       =( 0.4g) / (50g/mol)                                      n = 0.008 moles         mass of  B = 0.6g              so moles presentin B = mass / MW                                       =( 0.6g) / (80g/mol)                                      n= 0.0075 moles     From the balenced equation onemol A react with one mol B produce one mol C. So B is the limiting reagent . since itwas completely consumed in the reaction . b) we know per one mol of Breact with one mol A produce one mol C .                    Now here 0.0075 mol of B react with 0.0075 mol of A and produce0.0075 mol of C               so theoretical yield of C = (1.00of C /1.00mol ofB)*0.0075 mol of B                                                     =0.0075 mol of C                                                      =0.0075mol of C *(98g / 1.00 mol of C)                                                     = 0.735 g     c) if 0.6 g of C willget                                                                                        the percent yield of the reaction = (mass actual / masstheoretical)*100%                                                                                                =(0.6 g /0.735)*100%                                                                                               = 81.63%          MW 50 g/mol        80g/mol                    98 g/mol     a) mass of A = 0.4 g              so moles presentin A = mass / MW                                       =( 0.4g) / (50g/mol)                                      n = 0.008 moles         mass of  B = 0.6g              so moles presentin B = mass / MW                                       =( 0.6g) / (80g/mol)                                      n= 0.0075 moles     From the balenced equation onemol A react with one mol B produce one mol C. So B is the limiting reagent . since itwas completely consumed in the reaction . b) we know per one mol of Breact with one mol A produce one mol C .                    Now here 0.0075 mol of B react with 0.0075 mol of A and produce0.0075 mol of C               so theoretical yield of C = (1.00of C /1.00mol ofB)*0.0075 mol of B                                                     =0.0075 mol of C                                                      =0.0075mol of C *(98g / 1.00 mol of C)                                                     = 0.735 g     c) if 0.6 g of C willget                                                        so moles presentin B = mass / MW                                       =( 0.6g) / (80g/mol)                                      n= 0.0075 moles     From the balenced equation onemol A react with one mol B produce one mol C. So B is the limiting reagent . since itwas completely consumed in the reaction . b) we know per one mol of Breact with one mol A produce one mol C .                    Now here 0.0075 mol of B react with 0.0075 mol of A and produce0.0075 mol of C               so theoretical yield of C = (1.00of C /1.00mol ofB)*0.0075 mol of B                                                     =0.0075 mol of C                                                      =0.0075mol of C *(98g / 1.00 mol of C)                                                     = 0.735 g     c) if 0.6 g of C willget                                                                                        the percent yield of the reaction = (mass actual / masstheoretical)*100%                                                                                                =(0.6 g /0.735)*100%                                                                                               = 81.63%

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