Question

If a nylon 6.6, having a number average mas Mn=3.395 kg mol-1, was produced by RA2 + RB2 step polymerization with H2O being the by-product, determine the percent conversion if 1.1 moles of RA2 and 1 mole of RB2 were used in the process

Answer 1

Answer: 1. Actually nylon 6.6 is prepared by the reachion Oly two monomer ºn the conda Solion pacccess. ij Henta methylene diamic (NH12 - (CH) 6 - N119) fis Adipic acid (Coop - (E119), – COOH NH2 – CCH2)6 – NH2 + COOH – (CH2)4 - COOH. HI + (-HN – CCH2)6 - Noo c - CCH2),- Colo- 0 +1130 coepend ond) 8. Avarage molecular weight of nylon 6.6 is m= 226 zglmol [form this repeat unit] Mo= 113g Imol. Now we know that number avaroge molecular weight mo= In mo where, In = number average degree of polymarization Mo - avaroge. molecular weight of - Nylon 6.6 = 1139 Imol. Min = number avaroge moleculer Weight = 3.395 kg Imol. = 3.395*10*gImol {: 1 kg = 10$j]

i from w we get In Mn 3:395*103 g mol Mo 113 9 Imol Zn = 30.04. → avaroge degree of polymerization Now the number avarage degree of poly and Entent of reaction delate us. in ito - > 147-23P Where., po Exterit of reaction 7 = Starchi metric impalance bactor. - mole ob monomat 1 deacted mole of nono mar 2 veacted mole oly monomad RAZ mole of monoma, RB2 1 from cu) we get. 30.04 - 1+ 1+1.1-2x 1.17 → 30.04 = 21-2.32 > 63,084 -66.088p - 21 → + 66.088 P = +60.984 > P=0.9222 P=0.9227 X100% → P=92,27% :Percenil Conversation is 92027.