Question

Se supun monoclorurarii catalitice 97,5L benzen (p=0,88g/cm3)
a) Calculati volumul de clor (masurat in conditii normale) necesar.
b) Calculati procentul masic de clor in monoclorobenzen.

Answer 1

85,8                                            y

C6H6(sau nucleu benzenic) +Cl2→ C6H5-Cl+ HCl

     78                                        22,4

 

 

Pe sageata trebuie sa pui FeCl3

ρ=m/V

m=ρxV

m=97,5x1000x0,88

m=85,8x1000

m=85800 g benzen=85,8 Kg

y=22,4x85,8/78

y=24,64 mcubi

b) M=6x12+5+35,5

M=112,5 g/mol

112,5…………………..35,5

100………………………..t

 

T=100x35,5/112,5

T=31,55% Cl