Question

This is a dry lab assessment for synthesis of Nylon 6,6. I believe I understand how to calculate the limiting reagent (Q1) and have determined it to be adipic acid. I’m stuck on calculating the theoretical yield (Q2) and need some help getting through the bullet points of the procedure used to answer the question. Thanks!

1. In the experiment, you are instructed to use 0.500 g of adipic acid and 0.40 g of hexamethylenediamine. Calculate the mmol of each reagent that will be used in the experiment. Which reagent is the limiting reagent? Reagent Mol. Wt. Adipic acid 165 g/mol Hexamethylenediamine 116 g/mol liriting nergent 0.500 g adipic acid. acid. I mol adipic acid 1000 mmol adipic acid = 3.03 mmol adipic 165 g adipic doid I md adipic acid L acid 0.40 a hexamethylenediartine. Mol examecenediamine 1000 mmol hexamethylenediarine R4 mil Hog hexamethylenediarine Because reagents these I mol hexamethylenediamine hexamethylerediarine react in a 1:1 molar ratio, adipic acid (3.03 mmol) is the Cinciting reagent 2. What is the theoretical yield (mg) of nylon that is formed in the experiment? Since you do not know the molecular weight of the polymer, you can get the theoretical yield in using the following procedure: • Determine the mass of adipoyl chloride used from the mmol of the limiting reagent. Determine the mass of the hexamethylenediamine used from the mmol of the limiting reagent. • Determine the mass of H2O lost from the mmol of the limiting reagent. Theoretical yield of polymer = mass adipoyl chloride + mass hexamethylenediamine - mass HCI.

Answer 1

1. In the experiment, you are instructed to use 0.500 g of adipic acid and 0.40 g of hexamethylenediamine. Calculate the mmol of each reagent that will be used in the experiment which reagent is the limiting reagent? Reagent Mol. Wt. Adipic acid 165 g/mol Hexamethylenediamine 116 g/mol 0 500 g adipic od adige dialogic acid to mmol adapi od 15 g piedod maladii acid landing regest 3.03 mmol idigie sed 40 hesauri ol xanda 1000 tree Because there a r e s o anchalinidae act in a 11 molar raro, dipe acid (3.03 ) is the inciting reagert 2. What is the theoretical yield (mg) of nylon that is formed in the experiment? Since you do not know the molecular weight of the polymer, you can get the theoretical yield in using the following procedure: • Determine the mass of adipoyl chloride used from the mmol of the limiting reagent Determine the mass of the hexamethylenediamine used from the mmol of the limiting reagent. So you had calculated that the limiting reagent • is adific acid. (2 m.mol of Aclific acid used = 3.03 m mal af As we know, no. of immal will be produced R of adipoyl chloride to equal to Aditric acid ic. 1. Limiting Regent now, mmel of Adipoyl chloride = inmol of Adific aud. Mass of Adipoyl Chl. = 3.03% 103 mol Mol wit of Adipoge then Mass of Adipoyl Chlo. = 3.03 x 10 mol 183,03 g malt | Mass of Adipoyl chloride = 0.55g] • Determine the mass of H.O lost from the mmol of the limiting reagent Theoretical yield of polymer mass adipoyl chloride + mass hexamethylendiamine mass HCI. -3

b) Mass of HMD A (Henamithylene diombinu) word = ? As we know, HM DA and Alithic acid are reacting in 1:1 molar ratio. So, same na of males of HM DA will be used as of Adific acid modes of HMDA = moles of Adific and Mass of HM DA = 3:03 x 103 mal Mol. Wt. of HM DA Mass of HM DA = 3.038 10 mol 0 2 -3 116 g molt I Mass of HMD A=0.35 g)

y mass of the = 3.038103 mol Mass of the lost Moly of 40 = > Mokes of Adific and 189 med Mass Massotho = in 0. Mund 40 005) 3:03 *o*mel 3.03 X10 mol. Mol.ut of the d Theoritical field of polymer = ? Mass of HU = Males of Adipic acid Molist of the Mass of the = 3.03*103 x 36.5 Mass of the = o.lig.] Theoritical yield = Mass of Adipol chloride + Mass of HMDA = 0.55 + 0.35 – 0.11 Theoritical yield = 0,79g - Mass of the