Question

la clorurarea catalitica a toluenului se obtine o masa de reactie de 1391,5 grame formata din 36,36% o-clortoluen ,27,27% p-clortoluen ,23,14% 2,4-diclortoluen si 13,22% toluen nereactionat (procente de masa).Stiind ca amestecul reactant initial contine 13,75 moli clor ,randamentul de transformare a clorului este:
a)75%
b)85%
c)80%
d)82,5%

Answer 1

1391,5g amestec.............100%
xg orto...........................36,36%
yg para..........................27,27%
zg dicloro......................23,14%
tg nereact......................13,22%
x=506g ortoclorotoluen
y=379,5g paraclorotoluen
z=322g diclorotoluen
t=184g toluen nereactionat
toluenul nereactionat nu-l voi mai scrie
............1mol........126,5g
C7H8 + Cl2 -> o-C7H7Cl + HCl
..........a=4moli......506g
.............1mol......126,5g
C7H8 + Cl2 -> p-C7H7Cl + HCl
..........b=3moli.....379,5g
.............2moli......161g
C7H8 + 2Cl2 -> C7H6Cl2 + 2HCl
...........c=4moli....322g
din reactii => 11moli
randamentul=11*100/13,75=80%