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===========> K
NO3
-3 (
N+ 3= 0)=======> +5 [(1+
N+(-6)=0]
=========== +8 ==========> oxidze
.. ... ..... . .. .. ... .. .. .................. ..........................
then for K
MnO4
K
Mno4 ===========>
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