roxie2love

10+ Year Member
Dec 16, 2008
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Pre-Dental
I can't seem to balance this equation properly, any help is appreciated! Thanks in advance!

KMnO_4 + NH_3 -> KNO_3 + MnO_2 + KOH + H_2_O



Answer : 8KMnO_4 + 3NH_3 -> 3KNO_3 + 8MnO_2 + 5KOH + 2H_2_O
 
Last edited:

doc3232

10+ Year Member
7+ Year Member
Feb 15, 2008
3,809
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Status
Dental Student
I can't seem to balance this equation properly, any help is appreciated! Thanks in advance!

KMnO_4 + NH_3 -> KNO_3 + MnO_2 + KOH + H_2_O



Answer : 8KMnO_4 + 3NH_3 -> 3KNO_3 + 2MnO_2 + 5KOH + 2H_2_O
That's tough. IMO, it is rather time consuming also. I don't think it would be on the test.
 

Danny289

Member
10+ Year Member
Dec 2, 2006
1,523
4
Status
Pre-Dental
I can't seem to balance this equation properly, any help is appreciated! Thanks in advance!

KMnO_4 + NH_3 -> KNO_3 + MnO_2 + KOH + H_2_O



Answer : 8KMnO_4 + 3NH_3 -> 3KNO_3 + 2MnO_2 + 5KOH + 2H_2_O
KMno4=======> Mno2
Mn+7 +3e=======> Mn+4 3=reduction
-----------------------------------------------
NH3=======> KNO3
N-3 ======> N+5 +8e 8= oxidation
now switch 8 and 3 in your equation and....
------------------------------
recall from G chem
Kmno4 ==> finding oxide number for Mn ==> 1+ X +(-2. 4)=0===> X=+7
for NH3===> finding for N===> X+3=0 ===> X= -3 and......................
 
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