Balance this equation

[SMC2UCLA2_]

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KMnO4 + NH3 --> KNO3 + MnO2 + KOH + H2O

This is from a kaplan practice test and I am having a real hard time balancing this equation. Perhaps I am missing something.

Im looking for tips and tricks on how to balance this because I do have the final answer but it doesn't help me understand how to balance this one. I don't usually have a hard time with these but this one is a royal pain in the ass.

 

[tinman831]

KMnO4 + NH3 --> KNO3 + MnO2 + KOH + H2O

This is from a kaplan practice test and I am having a real hard time balancing this equation. Perhaps I am missing something.

Im looking for tips and tricks on how to balance this because I do have the final answer but it doesn't help me understand how to balance this one. I don't usually have a hard time with these but this one is a royal pain in the ass.

seems like there is a misprint somewhere in the initial equation

for every 1K MnO4 you have, you need 1 KNO3 and 1 MnO2 to keep the K and Mn balanced, however, the Oxygens would not equal.

 

[SMC2UCLA2_]

seems like there is a misprint somewhere in the initial equation

for every 1K MnO4 you have, you need 1 KNO3 and 1 MnO2 to keep the K and Mn balanced, however, the Oxygens would not equal.

Heres the answer:

8KMnO4 + 3NH3 --> 3KNO3 + 8MnO2 + 5KOH + 2H2O

and it seems to be balanced.

 

[tinman831]

Heres the answer:

8KMnO4 + 3NH3 --> 3KNO3 + 8MnO2 + 5KOH + 2H2O

and it seems to be balanced.

oops... thats what i get for trying to work on a problem at 3am in the morning 😳

 

[Richnator]

KMnO4 + NH3 --> KNO3 + MnO2 + KOH + H2O

This is from a kaplan practice test and I am having a real hard time balancing this equation. Perhaps I am missing something.

Im looking for tips and tricks on how to balance this because I do have the final answer but it doesn't help me understand how to balance this one. I don't usually have a hard time with these but this one is a royal pain in the ass.

Ok, I split the reaction into its oxidation and reduction components.
Reduction:
MnO4- + 4H+ + 3e- ===> MnO2 + 2H20

Oxidation:

NH3+ 3H20 ===> NO3- + 9H+ + 8e-

I eliminated the K+ cation because it is a specator ion. I then used the least common mutliple of both 3 and 8, which is 24. Therefore, I multiplied the reduction part by 8 and the oxidation part by 3.

Reduction:
8MnO4- + 32H+ + 24e- ===> 8MnO2 + 16H20

Oxidation:
3NH3+ 9H20 ===> 3NO3- + 27H+ + 24e-

Adding the two reduction and oxidation parts together, I cancel out the electrions and get:

In Acidic environment:
8MnO4- + 32H+ + 3NH3 + 9H20 ===> 8MnO2 + 16H20 + 3NO3- + 27H+

Since the top equation had OH-: I must add 32 OH- to both sides to cancel out the H+

Next, eliminate the surplus H20s
Balanced Reaction:
8MnO4- + 3NH3 =====> 8MnO2 +2H20 + 3NO3- + 5OH-