Zadatak ;)

Submitted by sp4soje on Sun, 12/21/2014 - 12:04
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U 39g benzena u prisustvu FeCl39katalizator0,doda se 1mol broma.Izracunajte mase supstanci nakon zavrsetka reakcije.

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milicaradenkovic - Mon, 12/22/2014 - 08:20

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