Question

1) Toluenul este clorurat fotochimic, obtinandu-se doua substante A si B. Substanta A are compozitia procentuala masica 44,1 % clor, 52,1% carbon si restul hidrogen. Substanta B are raportul masic C:H:Cl = 16,8:1:21,3. Determina:
a) formulele compusilor Asi B;
b) volumul de clor (c.n.) necesar obtinerii compusilor A si B, daca masa de toluen consumata este de 368 kg, iar compusii A si B se afla in raport molar 3:1.

Answer 1

=> %H=3,8
………………………………………|÷1,2422
C: 52,1/12=4,3416...........3,5
H: 3,8/1=3,8.....................3
Cl: 44,1/35,5=1,2422…....1
f bruta: (C3,5H3Cl)n
deoarece toluenul are 7 atomi de C => n=2
A: C7H6Cl2 - clorura de benziliden
………………………………..|÷0,6
C: 16,8/12=1,4………….2,33
H: 1/1=1.....................1,66
Cl: 21,3/35,5=0,6........1
f.bruta: (C2,33H1,66Cl)n
toluen - 7 atomi de C => n=3
B: C7H5Cl3 - clorura de benzin
b)
n=m/M => ntoluen=mtoluen/Mtoluen=368/92=4kmoli
3:1 => 3x kmoli A si x kmoli B
1kmol...44,8m³...1kmol
C7H8 + 2Cl2 -> C7H6Cl2 + 2HCl
3x.........134,4x.....3x
1kmol...67,2m³...1kmol
C7H8 + 3Cl2 -> C7H5Cl3 + 3HCl
x kmoli..67,2x.....x kmoli
4x=4 => x=1kmol
VCl2=201,6x=201,6m³ Cl2