M.C4H10 = 58 kg/kmol
=> 210/58 = 3,62 kmoli C4H10
a kmoli a a
C4H10 + Cl2 --500oC--> C4H9Cl + HCl , 1-cloro
1 1 1
b kmoli b b
C4H10 + Cl2 --500oC--> C4H9Cl + HCl , 2-cloro
1 1 1
c kmoli c
C4H10 --------------------------> C4H10 netransformat
1 1
c% = mdx100/ms
=> md.NaOH = c%xms/100 = 400x32/100 = 128 kg
128 kg n kmoli
NaOH + HCl --> NaCl + H2O
40 1
=> n = 128x1/40 = 3,2 kmoli HCl
a+b+c = 3,62
a+b = 3,2
-------------------
c = 0,42 kmoli C4H10 netransformati
=> m.C4H10.netransf. = 0,42x58 = 24,36 kg
M.1-cloro sau 2-cloro = 92,5 kg/kmol
masa de derivati este 3,2x92,5 = 296 kg
100% amestec derivati ........ 21% 1-cloro ........... 79% 2-cloro
296 kg .................................... x kg ...................... y kg
x = 62,16 kg , y = 233,84 kg
