Question

O arena mononucleara A,care contine 8,69%H,formeaza prin clorurare la lumina un compus clorurat B care contine 44,1%Cl.Determinati:
A.formulele moleculare ale compusilor A si B.
B.cantitatea de compus B care rezulta din 41,4g hodrocarbura A, rabdamentul reactiei fiind de 80%

Answer 1

Formula generală a hidrocarburilor aromatice mononucleare este CnH2n-6
%H=mH/M*100=>8,69=2n-6/14n-6*100=>8,69(14n-6)=200n-600=>121,66n-52,14=200n-600=>78,34n=547,86=>n=7
Formula compusului A este C7H8(toluen)
C7H8+2h-cl => C7H6Cl2 + 2HCl. Formula compusului B este C7H6Cl2.
41,4gC7H8..........161gC7H6Cl2
92gC7H8............x
x=92*161/41,4=>x=357.77g C7H6Cl2
randamentul=mp/mt*100=>mp=randament*mt/100=80*357,77/100=286.21g sau mt=mp*100/randament=357,77*100/80=447.21g C7H6Cl2