Rastvori

Submitted by Srdjan95 on Tue, 07/07/2015 - 18:25
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Izracunati masu(i kolicinu) dobijene soli  mesanjem 400 g rastvora NaOH masenog udela,u %,20% i 243,33g rastvora AlCl3 masenog udela,u %,30%

resenje:42,362 g i 0,359 mol Na[Al(OH)4]

Srdjan95's picture

Srdjan95 - Tue, 07/07/2015 - 18:31

I treba mi procentna koncentracija rastvora ove iste soli

resenje:6,58%

buduci_student's picture

buduci_student - Tue, 07/07/2015 - 18:54

AlCl3 + 4 NaOH = Na[Al(OH)4] + 3 NaCl

Preko ove jednacine uradis ovaj zadatak

 

 

Srdjan95's picture

Srdjan95 - Tue, 07/07/2015 - 18:56

ok

buduci_student's picture

buduci_student - Tue, 07/07/2015 - 22:01

Da li si uspio da uradis ovaj zadatak?

Srdjan95's picture

Srdjan95 - Tue, 07/07/2015 - 22:29

ne verujem da bi znao,probaj ti

buduci_student's picture

buduci_student - Tue, 07/07/2015 - 23:24

w1=0,2

mr=400

ms=mr*w=400*0,2=80g

n(NaOH)=80/40 =2mol

w2=0,3

mr2=243,33g

ms2=mr2*w2=243,33*0,3=73g

n=73/133,5 =0,5468mol

n(NaCl)=3*n(AlCl3)=3*0,5468mol =1,6404mol

n(NaOH)-n(NaCl)=2-1,6404mol =0,3596 mol n( natrijum tetrahidroksoaluminata)

n=m/M

m=n*M=0,3596*118=42,4328

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Srdjan95 - Tue, 07/07/2015 - 23:29

jos jednom hvala