Analiticka hemija

Submitted by Sofija333 on Mon, 06/26/2017 - 11:28
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Izracunati masu (i kolicinu) dobijene soli mesanjem 400 g rastvora NaOH masenog udela, u %, 20% i 243,33 g rastvora aluminijum-hlorida masenog udela , u %, 30%.

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milicaradenkovic - Thu, 06/29/2017 - 12:15

3NaOH+AlCl3------------------------->3NaCl+Al(OH)3

ms=w*mr=400*0.2=80 g NaOH

ms=mr*w=243.33*0.3=73 g AlCl3

n(baze)=ms/Mr=80/40=2 mol

n(AlCl3)=ms/Mr=73/133.5=0.5468 mol

u visku je baza pa su molovi soli n(NaCl)=3*n(AlCl3)=3*0.5468=1.64 mol

m(NaCl)=n(NaCl)*Mr=1.64*58.5=95.94 g