Question

27 28 oare cum se rezolva ?

Answer 1

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