alex1231

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Oct 24, 2006
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Hi
Does anyone know how to balance this?
KMnO4+NH3 goes into KNO3+MnO2+KOH+H2O
Thank you
 

alex1231

10+ Year Member
7+ Year Member
Oct 24, 2006
16
0
Status
  1. Pre-Dental
Here is the balanced equation but I have no idea how to get that!
8 KMnO4 + 3 NH3 → 3 KNO3 + 8 MnO2 + 5 KOH + 2 H2O
 
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Danny289

Member
10+ Year Member
Dec 2, 2006
1,523
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Status
  1. Pre-Dental
Hi
Does anyone know how to balance this?
KMnO4+NH3 goes into KNO3+MnO2+KOH+H2O
Thank you

Hi, alex, just follow my explanation and learn it forevere.

first: find out which elements reduce or oxidize,
you can easyly will find "N" and "Mn" why not others?
answer: K , Na , Li, when they are in compond always are"+1"
ok now we find oxidation number:


NH3===========> KNO3

-3 (N+ 3= 0)=======> +5 [(1+N+(-6)=0]

=========== +8 ==========> oxidze

.. ... ..... . .. .. ... .. .. .................. ..........................

then for KMnO4


KMno4 ===========>Mn02

Mn+7=========>Mn+4

===== 3 ====== red0x


Ok now we are almost done:
1- change the numbers (8 and 3)

8 KmnO4 + 3 NH3==> KNO3 + MnO2 + KOH + H20

2- in your product balance Mn & N because they are your main oxidation and reduction agents

8 KmnO4 + 3 NH3==> 3 KNO3 + 8 MnO2 + KOH + H20


3 -you go for metal or cation and anion that are not changed (here means K)

8 KmnO4 + 3 NH3==> 3 KNO3 + 8 MnO2 +5 KOH + H20


4- remeber most of the time water and hydrogyn and oxygyn are the last for balancing:

8 KmnO4 +3NH3 ==> 3 KNO3 + 8 MnO2 +5 KOH + 2 H20

Hope it was helpfull
good luck
 
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