[Chem] question

  • Thread starter Thread starter 299678
  • Start date Start date
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
This equation cannot be balanced as written, this is because in order to balance the oxygen on the left to the right side, the KMnO4 compound must be increased, however for every increase of this molecule you must also increase KNO3 and MnO2 which together increase oxygen at a higher rate and make balancing impossible.
 
Break the equation into each of its two parts:

MnO4^- ---> MnO2
NH3 ---> NO3^-

Balance oxygens by adding H2O
Next, balance the hydrogens by adding H^+

MnO4^- + 4H^+ + 3e^- ---> MnO2 + 2H2O
NH3 + 3H2O ---> NO3^- + 9H^+ + 8e^-

Multiply the first equation by 8, and the second by 3;

8MnO4^- + 32H^+ + 24e^- ---> 8MnO2 + 16H2O
3NH3 + 9H2O ---> 3NO3^- + 27H^+ + 24e^-

Adding these two will give

8MnO4^- + 3 NH3 + 5H^+ ---> 8MnO2 + 3NO3^- + 7H2O

Now, add 5OH^- (since its acidic) to both sides, cancelling out the H2O

8MnO4^- + 3 NH3 + 5H2O ---> 8MnO2 + 3NO3^- + 7H2O + 5OH^-
8MnO4^- + 3 NH3 ---> 8MnO2 + 3NO3^- + 2H2O + 5OH^-

8KMNO4 + 3NH3 ---> 3KNO3 + 8MnO2 + 5KOH + 2H2O

I hope this helps..
 
I was wondering how to do that too because I saw that same question on a Kaplan test. Thanks casper
 
Break the equation into each of its two parts:

MnO4^- ---> MnO2
NH3 ---> NO3^-

Balance oxygens by adding H2O
Next, balance the hydrogens by adding H^+

MnO4^- + 4H^+ + 3e^- ---> MnO2 + 2H2O
NH3 + 3H2O ---> NO3^- + 9H^+ + 8e^-

Multiply the first equation by 8, and the second by 3;

8MnO4^- + 32H^+ + 24e^- ---> 8MnO2 + 16H2O
3NH3 + 9H2O ---> 3NO3^- + 27H^+ + 24e^-

Adding these two will give

8MnO4^- + 3 NH3 + 5H^+ ---> 8MnO2 + 3NO3^- + 7H2O

Now, add 5OH^- (since its acidic) to both sides, cancelling out the H2O

8MnO4^- + 3 NH3 + 5H2O ---> 8MnO2 + 3NO3^- + 7H2O + 5OH^-
8MnO4^- + 3 NH3 ---> 8MnO2 + 3NO3^- + 2H2O + 5OH^-

8KMNO4 + 3NH3 ---> 3KNO3 + 8MnO2 + 5KOH + 2H2O

I hope this helps..

Thanks you for the post.
 
Top