Question

se alchileaza 780 kg benzen cu etena. se obtine un amestec de etibenzen, dietilbenzen si benzen in raport molar de 5:0.5:1. calculati masa de etilbenzen care se obtine si randamentul procesului

Answer 1

n=m/M=780/78=10 kmoli C6H6

x/y=5/0,5 => y=0,5x/5

x/z=5/1 => z=x/5

x+y+z=10 => x+0,5x/5 + x/5 = 10 => 6,5x/5=10 => x=50/6,5 =>x=7,6923 kmoli C6H5-CH2-CH3

y=0,76923 kmoli C6H4-(CH2-CH3)2

z=1,53846 kmoli C6H6 nereactionat

m C6H5-CH2-CH3=7,6923*106=815,3838kg C8H10

n C6H6 reactionat = 10-1,53846 = 8,461151

n(eta) = mp/ mt *100 = 7,6923/8,461151*100 ≈ 90,911%