27. MA = 137kg/kmol
teoretic : 27400/80 = 342,5 kg A
n = m/M = 342,5/137 = 2,5 kmoli A
n A = nC6H5-NH2 = 2,5 kmoli pur
m pur = 2,5*93 = 232,5kg C6H5-NH2
m impur = 23250/80 = 290,625kg C6H5-NH2
28. teoretic: 13700/80 = 171,25kg A
MA = 137kg/kmol
n = m/M = 171,25/137 = 1,25 kmoli A
nA = noxid de etena = 1,25 kmoli
m oxid de etena = 1,25*44 = 55kg
100%..........................55kg
110%...........................x = 60,5kg C2H4O
