Stehiometrija

Submitted by Alexstrasuz on Fri, 02/13/2015 - 23:52
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Koliko mililitara 20% rastvora HCl gustoce 1.1g/ml treba za neutralizaciju 45g 15% rastvora KOH

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milicaradenkovic - Sat, 02/14/2015 - 07:41

HCl+NaOH--------------------->NaCl+H2O

ms=mr*w=45*0.15=6.75 g

n=ms/Mr=6.75/40=0.16875 mol

n(HCl)=n(NaOH)=0.16875 mol

ms=n(HCl)*Mr=36.5*0.16875=6.16 g

mr=ms/w=6.16/0.20=30.8 g

V(HCl)=mr/ro=30.8/1.1=28 ml

 

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Alexstrasuz - Sat, 02/14/2015 - 08:10

I ja dobijam to rijesenje dok u  zbirci je 100ml

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milicaradenkovic - Sat, 02/14/2015 - 18:26

Izvini greska je u jednacini nije NaOH nego KOH.

n=ms/Mr=6.75/56=0.1205 mol

n(HCl)=n(KOH)=0.1205 mol

ms=n*Mr=36.5*0.1205=4.398 g

mr=ms/w=4.398/0.2=22 g

V(HCl)=mr/ro=22/1.1=20 ml nije dobro resenje.