Chlorobenzene (C6H5Cl), an important solvent and intermediate in the production

Question

Chlorobenzene (C6H5Cl), an important solvent and intermediate in the production of many other chemicals, is produced by bubbling chlorine gas through liquid benzene in the presence of ferric chloride catalyst. In an undesired side reaction, the product is further chlorinated to dichlorobenzene, and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene.

The feed to a chlorination reactor consists of essentially pure benzene and a technical grade of chlorine gas (98 wt% Cl2, the balance gaseous impurities with an average molecular weight of 25.0). The liquid output from the reactor contains 65.0 wt% C6H6, 32.0% C6H5Cl, 2.5% C6H4Cl2, and 0.5% C6H3Cl3. The gaseous output contains only HCl and the impurities that entered with the chlorine.

Find
% excess Benzene
Fractional conversion of benzene
Fractional yield of monochlorobenzene
Mass ration of gas feed to the liquid feed

PLEASE USE EXTENT METHOD

Explanation / Answer

The reactions are C6H6+ Cl2-----àC6H5Cl+ HCl

C6H5Cl + Cl2----àC6H4Cl2+HCl

C6H4Cl2+Cl2----àC6H3Cl3+HCl

Basis : 100 kg of liquid product

It contains 65 kg Benzene, 32kg of C6H5Cl, 2.5 kg C6H4Cl2, 0.5 kg C6H3Cl3

Molar masses : C6H6= 78, C6H5Cl= 112.5, C6H4Cl2=147, C6H3Cl3= 181.5

Moles of trichlorobenzene in 0.5 kg =0.5/181.5= 0.002755

Moles of dichlorobenzene required for formation of trichlorobenzene=0.002755

Moles of monochlorobenene required =0.02755 and Benzene required= 0.002755

Similarly, moles of dichlorobenzene= 2.5/147=0.017

Moles of Benzene required =0.017

Total moles of Benzene for formation of monochlorobenzene+ dichlro benzene + trichlorobenzene = 0.017+0.02755+32/112.5= 0.33

Moles of Chlorine required= 0.33

Chlorine supplied is 98% pure, chlorine supplied= 0.33/0.98= 0.34 moles

Molar mass of 98% Chlorine containing other gases whose molecular weight is 25

=0.98*71+0.02*25= 70 kg/kgmoles

Mass of chlorine supplied= 0.34*70.8= 24 kg

Mass of Benzene reacted = 0.33*78= 25.74 kg

Mass of Benzene unreacted= 65 kg

Total Benzene supplied= 65+25.74= 90.74 kg

Benzene consumed to form monochlorobenzene=(32/112.5)*78= 22.2 kg

Excess benzene= 90.74-22.20= 68.54 kg

% excess benzene =100*68.54/22.20 =308%

Benzene converted=25,74 kg , Benzene converted= 25.74/90.74= 0.283

Fractional yield of Monochlorobenzene = 22.2/90.74=0.244

Mass ratio of gas feed (cl2) to benzene (liquid feed)= 24/90.74=0.264

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