- Joined
- May 8, 2012
- Messages
- 36
- Reaction score
- 3
The reaction unbalanced reaction is:
KMnO4 + NH3 ---> KNO3 + MnO2 + KOH + H2O
every time I balance it using the half equation method I get:
3(NH3) + 8 (KMnO4) ---> 3 (KNO3) + 8 (MnO2) + 5 (K+) + 7 (OH-) + 4 (H2O)
That is incorrect however,
I am assuming this reaction is being done in basic solution (KOH) so I am adding OH to balance the oxygens and H2O to balance the hydrogens? Any ideas on what I am doing wrong? How do I solve this redox to get the right amount of water and hydroxide molecules?
KMnO4 + NH3 ---> KNO3 + MnO2 + KOH + H2O
every time I balance it using the half equation method I get:
3(NH3) + 8 (KMnO4) ---> 3 (KNO3) + 8 (MnO2) + 5 (K+) + 7 (OH-) + 4 (H2O)
That is incorrect however,
I am assuming this reaction is being done in basic solution (KOH) so I am adding OH to balance the oxygens and H2O to balance the hydrogens? Any ideas on what I am doing wrong? How do I solve this redox to get the right amount of water and hydroxide molecules?