Smeša KCl i NaCl

Submitted by bojank11 on Sat, 11/30/2013 - 18:43
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Iz uzorka smese KCl i NaCl ukupne mase 0.1225 g dobijen je talog AgCl cija je masa 0.2850 g. Izracunati procentni sastav KCl i AgCl u smesi.

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milicaradenkovic - Sat, 11/30/2013 - 18:59

m(smese)=0.1225 g

m(AgCl)=0.2850 g

n(AgCl)=m(AgCl)/Mr=0.2850/143.5=0.00199 mol

n(NaCl)=n(AgCl)=0.00199 mol

m(NaCl)=n(NaCl)*Mr=0.00198*58.5=0.1164 g

w(NaCl)=m(NaCl)/m(smese)=0.1164/0.1225=0.95=95 %

w(KCl)=100-w(NaCl)=100-95=5 % da nisi ti mislio procenat KCl i NaCl u smesi?

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bojank11 - Sat, 11/30/2013 - 19:05

KCl i NaCl... greska...

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bojank11 - Sat, 11/30/2013 - 19:09

treba maseni udeo KCl i NaCl u smesi

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milicaradenkovic - Sat, 11/30/2013 - 19:11

Resila sam ti zadatak .