Question

Prin clorurarea fotochimica a toluenului rezulta un amestec de clorura de benzil,clorura de benziliden si toluen in raport molar 5:2:1,5.aflati:
a.continutul procentual de clor al amestecului
b.masa de toluen luata in lucru,daca prin separarea HCl in apa se obtin 1,2 l solutie 2M la un randament de 80%

Answer 1

Cm=n/Vs => n=Cm*Vs=1,2*2=2,4moli HCl
80/100=npractici/nteoretici => nteoretici=2,4*100/80=3moli HCl
separarea se face cu r=80%, si nu clorurarea.
raport de 5:2:1,5 => 5x moli clorura de benzil, 2x moli clorura de benziliden, 1,5x moli toluen nereactionat
92g..............................1mol...........1mol
C6H5-CH3 + Cl2 -> C6H5-CH2Cl + HCl
92*5x..........................5x..................5x
92x...............................1mol.............2moli
C6H5-CH3 + 2Cl2 -> C6H5-CHCl2 + 2HCl
92*2x...........................2x....................4x
92g.........1mol
C7H8 -> C7H8
92*1,5x...1,5x
nHCl=9x=3 => x=0,33333moli
mtoluen=92(5+2+1,5x)=92*8,5*0,333=260,66g toluen
mbenzil=Mbenzil*5*0,333=210,83333g
mbenziliden=161*2*0,333=107,3333g
mtoluennereact=92*1,5*0,333=46g
mamestec=364,166g
mCl=35,5*(2*2x+5x)=35,5*3=106,5g
%Cl=106,5*100/364,166=29,24% Cl