Question

La alchilarea metilaminei cu clorura de metil se obtine un amestec de reactie care contine metilamina nereactionata ,dimetilamina si trimetilamina in raport molar 2:4:1
a)scrieti ecuatiile reactiilor corespunzatoare
b)calculati masa de dimetilamina obtinuta din21 moli metilamina
c) calculati conversia utila ,conversia totala si randamentul reactiei

Answer 1

4xa,mol         <-------------------------4x a,mol------------------------> 4x3mol
CH3-NH2 + CH3Cl---> CH3-N(CH3)H+HCL
1xa,mol          <------------------------1xa,mol---------------------------> 1x3mol
CH3-NH2+ 2CH3Cl----> CH3-N(CH3)2+ 2HCl
2xa,mol          <---------------------------2xa,mol----------------------> 2x3mol
CH3-NH2----------------------------------->CH3NH2
7a=21--. > a=3

Cu= 12x100/21=
Ct= (12+3)x100/21=
eta= Cu/Ct=                                       calculeaza !!!!!

Answer 2

n1 moli CH3-NH2 nereactionata
n2 moli CH3-NH2 + CH3-Cl ⇒ (CH3)2NH +HCl 
n3 moli CH3-NH2 + 2CH3Cl ⇒(CH3)3N + 2HCl 
n1/2 = n2/4 = n3/1 = 21/7 = 3    n1 = 6 moli;  n2 = 12moli ; n3 = 3moli 
n1 + n2 + n3 = 21
b) n dimetilamina = n2 = 12 moli 
c) Cu = 1200/21 = 57,14% 
Ct = (n2+n3)·100/n = 1500/21= 71,43% 
η = 57,14·100/71,45 = 79,97%≈ 80%