zadataka,analiticka hemija

Submitted by Student1337 on Tue, 06/17/2014 - 13:57
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Izracunati maseni udi hlora u uzorku mase 0,2491 g, ako je talozenje vrseno sa AgNO3 i dobijeni talog AgCl mase 0,2183 g. M(Cl) = 35,5 ; M(AgCl) = 143,4.

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milicaradenkovic - Tue, 06/17/2014 - 17:01

AgNO3+HCl--------------->AgCl+HNO3

n(AgCl)=m/Mr=0.2183/143.4=0.00152 mol

n(Cl)=n(AgCl)=0.00152 mol

m(Cl)=n(Cl)*Mr=0.00152*35.5=0.05396 g

w(Cl)=m(Cl)/m(uzorka)=0.05396/0.2491=0.2166=21.66 %

 

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Student1337 - Tue, 06/17/2014 - 17:40

hvala :)