Author Topic: about methcathinone  (Read 2569 times)

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skanic

  • Guest
about methcathinone
« on: November 03, 2001, 01:39:00 PM »
in the cat faq 2.2, it told about oxydising pseudoephedrine with kmno4.
But i would use pseudo HCl, then, i calculated :
  3gm(that's what i use)/201,69(mw of pseudo HCl) =0,0148mol(of pseudo HCl)
  0,0148/2,5 (2,5 mole of kmno4 oxydise 1 mole of pseudo)=5,9E-3
  5,9E-3*158(MW of KMnO4)=0,94g !!!
It means that i should use 0,94 g of KMnO4 if i use pseudo HCl instead of 1,148g if i use pseudoephedrine base...
please, tell me if i'm wrong because this could be the reasons of my nightmares.     
     
               skanic

Osmium

  • Guest
Re: about methcathinone
« Reply #1 on: November 03, 2001, 01:42:00 PM »
> 0,0148/2,5 (2,5 mole of kmno4 oxydise 1 mole of pseudo)=5,9E-3

That's wrong. Should be 0.0148 x 2.5

skanic

  • Guest
Re: about methcathinone
« Reply #2 on: November 04, 2001, 10:43:00 AM »
OH shit !
excuse me for that typo error, it sould be :
"1 mole of KMnO4 oxydises 2,5 moles of pseudo" ; But the problem stays the same and it's still
0,0148/2,5=5,9E-3
 
          Skanic

sorry for that error...