CnH2n+2 + xCl2-------> CnH2n+2-xClx + xHCl
M,derivat= 14n+2-x+35,5x=14n+2+34,5x
avand doua ecuatii trebuie alcatuit un sistem cu 2 ecuatii
M,derivat= 3,5oriM,CO2---> 3,5ori44
14n+2+34,5x= 154... .................................................o ecuatie
14n+2+34,5x g..............12n gC
100g.................................7,79 g
1200n=109n+15,58+ 268,75x........................a doua ecuartie
rezolva sistemul si vei obtine CCl4