Zadatak ;)

sp4soje's picture

U 39g benzena u prisustvu FeCl39katalizator0,doda se 1mol broma.Izracunajte mase supstanci nakon zavrsetka reakcije.

milicaradenkovic's picture

C6H6+Br2(FeCl3)------------------->C6H5Br+HBr

n(C6H6)=m/Mr=39/78=0.5 mol

n(C6H5Br)=n(C6H6)=0.5 mol jer je brom u visku

m(C6H5Br)=n(C6H5Br)*Mr=0.5*157=78.5 g

n(HBr)=n(C6H6)=0.5 mol

m(HBr)=n(HBr)*Mr=0.5*81=40.5 g