Răspuns :
reactiile le poti gasi intr-un manual de chimie de clasa X
b)
5a 5a
C6H5-CH3 --------------------> C6H5-CH3 netransformat
1 1
8a 8a
C6H5-CH3 + Cl2 --FeCl3--> Cl-C6H4-CH3 + HCl
1 1
4a 4a
C6H5-CH3 + 2Cl2 --FeCl3--> (Cl)2C6H3-CH3 + 2HCl
1 1
3a 3a
C6H5-CH3 + 3Cl2 --FeCl3--> (Cl)3C6H4-CH3 + 3HCl
1 1
--------------
m = 3680 kg toluen
=>
(5a+8a+4a+3a)x92 = 3680
=> a = 2 kmoli
=> m.monoclorotoluen = 8axM.monoclorotoluen
= 8x2x126,5 = 2024 kg monoclorotoluen
c)
Cu = (n.moli transf. in prod. util/n.moli totali introdusi)x100
= (8a/20a)x100 = 40%