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Author: Subject: Trying to seperate MnCl2 & KCl from dead reaction mixture
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[*] posted on 10-7-2012 at 05:19

So here is my proces in detail then but keep in mind I haven't done yet all the calculations but it's approx, will make better calculations when you guys think that the proces is okay.

2KMnO4 + 16HCl --> 8H2O + 2MnCl2 + 2KCl + (irrelevant 5Cl2 gas).

The MnCl2 will dissolve and most of the KCl also but there will be some left because the solution is satured so Ill add more water till everything is dissolved.

Then I add bleach saturated with NaOH. MnO2 formed, insoluble in water will precipate out will be filtered off and rinsed several times.
Now we will boil the mixture:

3NaClO --> 2NaCl + NaClO3 (because of boiling)

NaClO3 + KCl --> KClO3 + NaCl (changing ions)

HCl + NaOH --> NaCl + H2O (adding HCl)

Then all we gotta do is seperate the KClO3 from the NaCl which is done by putting it in the freezer and because of the lower solubility in water the KClO3 should come out of the solution while the NaCl stays in the solution.

So we have seperated MnO2 and KClO3 from the mixture, job done! A video about this will be made soon I will let you guys know thanks for helping!

[Edited on 10-7-2012 by CrEaTiVePyroScience]
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[*] posted on 11-7-2012 at 05:46

Or: you can buy MnO2 or MnSO4 cheap as chips! ;)

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