Question

Ex 42 plss!!! Punctul b !!!!!!!!!!!!

Answer 1

CH≡CH+ HCN---> CH2=CH-CN
n CH2=CH-CN-------->[-CH2-CHCN-]n

V=560mcub
n=V/Vm= 560mcub/22,4kmol/mcub= 25kmol acetilena impura
p= n,purx100/n,impur= > n,pur= 25kmolx70/100= 17,5kmol acetilena pura
-din ecuatiile scise, rezulta ca din 17,5kmol acetileana pura se obtin 17,5kmol polimer,daca randamentul ar fi 100%
-cu randament mai mic, se obtine mai putin
n,practic= 17,5kmolx80/100= 14 kmol polimer
-m= nxM= 14kmolx(36kgC+14kgN+3kgH)/mol=...................calculeaza !!

Answer 2

HC≡CH + HCN ⇒ H2C=CH-CN 
nH2C=CH-CN ⇒ -[-CH2-CH(CN)-]n- 
nC2H2 pur = 0,7·560/22,4 = 17,5kmoli = n monomer (teoretic) 
m monomer = 0,8·17,5·53 = 742Kg 
PNA = fibra sintetica, inlocuitor de lana