Kaplan Chem question help

[basarachobits]

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How do you attack this question?

Many thanks

 

[AskJeeves]

How do you attack this question?

Many thanks

What's the question???

 

[Baylor2011]

Also...ask this in the DAT Discussion forums - you'll get more answers

 

[khangreen]

Baylor's right, but I'll answer it anyways.

I think he's trying to balance it.

If so, then if you notice the nitrogen changes oxidation states from ammonia to nitrate, (-3 to +5 (H is +1 and O is -2)), you'll realize it's redox. Plus, manganese goes from +7 to +4. Separate them into half reactions and balance the elements/compounds/ions in those first, then balance for charge with electrons. Then, balance the number of electrons using the least common multiple and combine the equations. You can leave out the spectator ions out until the end to make it simpler. For the half reactions you should get the following:

3[e-] + [MnO(4)-] + 2[H(2)O] => [MnO(2)] + 4[OH-]
and
[NH(3)] + 9[OH-] => [NO(3)-] + 6[H(2)O] + 8[e-]

LCM is 24 so multiply the first half reaction by 8 and the second one by 3.
This gives you
24[e-] + 8[MnO(4)-] + 16[H(2)O] + 3[NH(3)] + 27[OH-] => 8[MnO(2)] + 32[OH-] + 3[NO(3)-] + 18[H(2)O] + 24[e-]

cancel things on both sides and you'll end up with

8[MnO(4)-] + 3[NH(3)] => 8[MnO(2)] + [3NO(3)-] + 5[OH-] + 2[H(2)O]

Put the spectator ions back in and you get

8 KMnO(4) + 3 NH(3) => 8 MnO(2) + 3 KNO(3) + 5 KOH + 2 H(2)O

Oh yeah, this question gave OH- somewhere in the equation, so Oxygen and Hydrogen were balanced with OH-. If they said to predict it using an acid of if they included the H+ instead, then you'd balance the Oxygen and Hydrogen with H+ (or H(3)O+).

 

[basarachobits]

Yes, I forgot to say this is a redox reaction problem.
Thanks a lot for solving it.

Baylor's right, but I'll answer it anyways.

I think he's trying to balance it.

If so, then if you notice the nitrogen changes oxidation states from ammonia to nitrate, (-3 to +5 (H is +1 and O is -2)), you'll realize it's redox. Plus, manganese goes from +7 to +4. Separate them into half reactions and balance the elements/compounds/ions in those first, then balance for charge with electrons. Then, balance the number of electrons using the least common multiple and combine the equations. You can leave out the spectator ions out until the end to make it simpler. For the half reactions you should get the following:

3[e-] + [MnO(4)-] + 2[H(2)O] => [MnO(2)] + 4[OH-]
and
[NH(3)] + 9[OH-] => [NO(3)-] + 6[H(2)O] + 8[e-]

LCM is 24 so multiply the first half reaction by 8 and the second one by 3.
This gives you
24[e-] + 8[MnO(4)-] + 16[H(2)O] + 3[NH(3)] + 27[OH-] => 8[MnO(2)] + 32[OH-] + 3[NO(3)-] + 18[H(2)O] + 24[e-]

cancel things on both sides and you'll end up with

8[MnO(4)-] + 3[NH(3)] => 8[MnO(2)] + [3NO(3)-] + 5[OH-] + 2[H(2)O]

Put the spectator ions back in and you get

8 KMnO(4) + 3 NH(3) => 8 MnO(2) + 3 KNO(3) + 5 KOH + 2 H(2)O

Oh yeah, this question gave OH- somewhere in the equation, so Oxygen and Hydrogen were balanced with OH-. If they said to predict it using an acid of if they included the H+ instead, then you'd balance the Oxygen and Hydrogen with H+ (or H(3)O+).