Good G'Chem Question

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dat_student

dat_student

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Balance the following rxn (show your steps):

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

I know, I know,...It's an easy question but try to see if you can balance this rxn in 54 seconds (100 sci. questions, 90 minutes: 54 seconds per question).
 
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dat_student said:
Balance the following rxn (show your steps):

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

I know, I know,...It's an easy question but try to see if you can balance this rxn in 54 seconds (100 sci. questions, 90 minutes: 54 seconds per question).
Click to expand...

Before everyone freaks out...there are PLENTY of easy questions on the DAT so you have more than 54 seconds for a question like this. 😉 Of course, you only have 38 seconds in DAT Achiever by the time you get to this question. 🙂 It's either dat achiever or top score, right dat_student?
 
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ok, let's do this: Can anyone balance that rxn in 540 seconds (i.e. 9 minutes)?

mocha: see my other post right above this post
 
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dat_student said:
Balance the following rxn (show your steps):

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

I know, I know,...It's an easy question but try to see if you can balance this rxn in 54 seconds (100 sci. questions, 90 minutes: 54 seconds per question).
Click to expand...
I think if it is under acidic environment, it would be:
2KMnO4 + NH3 + 2H+ + 2e- =========> KNO3 + MnO2 + KOH + 2H20
 
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Oh yah, you add H20 to balance the oxygen. H+ to balance the hydrogens and e- to balance the charges on both sides. If I am not wrong that is 🙂
 
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Richnator said:
I think if it is under acidic environment, it would be:
2KMnO4 + NH3 + 2H+ + 2e- =========> KNO3 + MnO2 + KOH + 2H20
Click to expand...

You should never end up having electrons in your balanced rxn. 😉
 
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dat_student said:
You should never end up having electrons in your balanced rxn. 😉
Click to expand...

U still didn't provide the answer for "Funny Math Question" so i bet this is also one of those question that has no answer, or doesn't even require a thought of more than 2 sec to get the answer, right?
 
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dat_student said:
You should never end up having electrons in your balanced rxn. 😉
Click to expand...
Ahh, yah. I realized my mistake. Forgot to perform the most crucial part. Split the reaction into the oxidation and reduction components.
 
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Flipper405 said:
What about nuclear reactions with beta particles? I thought those have to have electrons in them.
Click to expand...

not necessarily, gamma-decay, positron decay, alpha decay ddon't have electrons in them.
 
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Flipper405 said:
I just bumped that thread up again. 🙂 But yes, that's what I'm expecting. Maybe it'll just encourage me to figure it out for myself.
Click to expand...

I thought u gave up on that Funny Question even before I did. 🙂

how about I make a new funny question like....
if u fart under the water...what chemical reaction happens?
answer = H2O + fart oxygen = H2OOOOOOOOO....depends on how long u fart 🙂
 
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joonkimdds said:
I thought u gave up on that Funny Question even before I did. 🙂

how about I make a new funny question like....
if u fart under the water...what chemical reaction happens?
answer = H2O + fart oxygen = H2OOOOOOOOO....depends on how long u fart 🙂
Click to expand...

😱

I wouldn't go telling everyone that. :wow:
 
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dat_student said:
You should never end up having electrons in your balanced rxn. 😉
Click to expand...
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+

Balanced Reaction:
8MnO4- + 3NH3 + 14H20 ====> 8MnO2 +16H20 + 3NO3- + 5OH-
 
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dat_student said:
Balance the following rxn (show your steps):

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

I know, I know,...It's an easy question but try to see if you can balance this rxn in 54 seconds (100 sci. questions, 90 minutes: 54 seconds per question).
Click to expand...
I can believe I typed this whole entry up and it didnt post it. Anyway, to make it short.
This is the balanced rxn.

8MnO4- + 3NH3 +14H20 ====> 8MnO2 + 16H20 + 3NO3- + 5OH-

I initially did it as if it was a acid. Then added 27OH- to both sides and canceled the H20 to get this final equation. I also eliminted K+ because they were spectator Ions.
 
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Richnator said:
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+

Balanced Reaction:
8MnO4- + 3NH3 + 14H20 ====> 8MnO2 +16H20 + 3NO3- + 5OH-
Click to expand...


Good work.
 
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Flipper405 said:
Good work.
Click to expand...
Thanks, tookme 10 mins to do the problem and then type it up. I didn't want to leave that question unanswered. I also doubt they would give that in the DAT. Reason being it takes too much time to do.
 
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Richnator said:
I can believe I typed this whole entry up and it didnt post it. Anyway, to make it short.
This is the balanced rxn.

8MnO4- + 3NH3 +14H20 ====> 8MnO2 + 16H20 + 3NO3- + 5OH-

I initially did it as if it was a acid. Then added 27OH- to both sides and canceled the H20 to get this final equation. I also eliminted K+ because they were spectator Ions.
Click to expand...

not right. you should have the Ks in the final balanced rxn. This question has a solution 😉
 
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Mn+7 + 3e ---> Mn+4 ------ x8
N-3 -8e ---> N+5 ---------- x3

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