How To Balance Cu + HNO3 = Cu(NO3)2 + NO + H2O Watch

tesmifami
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I started my A-levels this year and we had a little test in chemistry to cover what we have done over the 4 or 5 weeks we have been at school and It was all fine expect for one question which I spent around 10 minutes on and still could not answer. The question was to balance this:

Cu + HNO3 = Cu(NO3)2 + NO + H2O

Since the exam I have looked up the solution on-line and discovered that the answer is:

3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O

My question is how would you go about working this out? I would never have worked this out and if I got a question like this in an exam I could spend ages on it and never figure it out. I have never struggled with balancing equations before but this one was so complex and hard I have no idea how you could work this out.

Obviously there must be some method, so could you tell me how you would go about balancing it. Or am I just being really slow and would I be expected to be able to look at that and solve it?

I guess in retrospect I could have looked at the H20 and the HNO3 and realised that the number before the HNO3 must be double the number before the H20 then done trail and error but that would still be time consuming.

Thanks for your help
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Substantia
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Dont worry you are very unlikely to get exam question that involves any multiple greater than 4.
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NutterFrutter
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(Original post by Substantia)
Dont worry you are very unlikely to get exam question that involves any multiple greater than 4.
Equations like this have came up before on Edexcel papers.

In redox reactions you can use oxidation numbers.
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charco
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(Original post by tesmifami)
I started my A-levels this year and we had a little test in chemistry to cover what we have done over the 4 or 5 weeks we have been at school and It was all fine expect for one question which I spent around 10 minutes on and still could not answer. The question was to balance this:
Cu + HNO3 = Cu(NO3)2 + NO + H2O
Since the exam I have looked up the solution on-line and discovered that the answer is:
3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O
It's not easy unless you understand the concept of oxidation states (these are the apparent valencies of the atoms within the species)

On the left hand side of the equation the copper is in the oxidation state zero (the element) and on the right hand side it is in the oxidation state +2. This means that it has to lose two electrons:

Cu --> Cu2+ + 2e

Meanwhile, the nitrogen atom on the left hand side is in the oxidation state +5. On the right hand side (in the species NO) it is in the oxidation state +2. Hence it must have gained three electrons. To remove the oxygen atoms you need hydrogen ions to make water.

NO3- + 4H+ + 3e --> NO + 2H2O

Now you have to combine the two equations by equalising the electrons. Multiply the first equation by 3 and the second by 2

3Cu --> 3Cu2+ + 6e
2NO3- + 8H+ + 6e --> 2NO + 4H2O
-------------------------------------------------------------
3Cu + 2NO3- + 8H+ --> 3Cu2+ + 2NO + 4H2O

Now we have to enter the spectator ions. You can combine two of the hydrogen atoms with the two nitrate ions to make nitric acid, but you need another six nitrate ions to combine with the six remaining hydrogen ions. The six nitrate ions are used to join up with the three copper(II) ions:

3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO + 4H2O
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fgha
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I had this quation too.I thought very much but I cant solve it.now I understand it by your explanaton.
Thanks for your perfect explanation .
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kevandmags2017
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You can solve this using simultaneous equations: You know the elements are Cu, H, N & O. Lets call the unknown coefficients a, b, c, d & e (write this full equation out yourself) Balancing each element in turn (which they must) yields 4 algebraic equations which will yield the ratio of the coefficients to each other.

1) Balancing for Cu gives a = c
2) ....for H gives b = 2e
3) ...for N gives b = 2c + d
4) ....for O gives 3b = 6c + d + e

if you multiply eq 3 by 3 you get 3b = 6c + 3d and joing with eq 4 yields 6c + 3d = 6c + d + e ie 3d = d + e ie 2d = e we now have 2 of the simple equations involving 3 of the coefficients so should be easy to solve from here: try out a value of d = 1 which => e = 2 and so eq 2 => b = 4. Eq 3 => 4 = 2c + 1 ie c = 3/2 and eq 1 => a = 3/2 so the ratio of the coefficients is 3/2:4:3/2:1:2 which is no good as it needs to be all integers and the minimum possible at that. Multiply the ratio throughout by 2 yields: 3:8:3:2:4
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seara
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I'snt the equation balanced this way: Cu 4HNO3 --> Cu(NO3)2 2NO2 2H2O
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seara
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I'snt the equation balanced this way:

Cu + 4HNO3 --> Cu(NO3)2 + 2NO2 +2H2O
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Batamiz
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Cu HNO3=Cu(NO3)2 NO2 2H20 BY HIT AND TRIAL
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Angel_Chen
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I don't do A level chemistry, I might next year, but looking at:

Cu + HNO3 = Cu(NO3)2 + NO + H2O
Left=L and Right=R
N.B. that HNO3 is the only important thing on L side.
I'd start by counting all oxygen as they are the most complicated.
L:3 and R:8, so balance... multiplying HNO3 by 8 and R has 3 compound with O so don't touch.
Now count L hydrogens, because we've only got 1 on each side with H atoms.
L:8 and R:2
multiply R by 4. now we have 8 H and 4 O
L:8 N and R:2+1 N
L:24 O and R:6+1+4
We know Cu(NO3)2 can only be 1, 2 or 3 so try and 3 is the only one that balances N and O.
Finally balance Cu on L
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Kavya Agarwal
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The answers given are wrong...today I asked for balancing a equation and it was all wrong Actually the equation was wrong balanced.....
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Kavya Agarwal
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Cu + HNO3= Cu (NO3)2 + NO + H2O
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Kavya Agarwal
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The answers given are wrong...today I asked for balancing a equation and it was all wrong Actually the equation was wrong balanced.....
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Shibu Ahmed
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If we are taking the only the reactants and products which involves in change in oxidation number, the given equation can be written as followsCu HNO3 --> Cu2 NOAfter balancing by using any one of the balancing method the above equation become 3Cu 2HNO3 6H --> 3Cu2 2NO 4H2OThis 6H will be from another 6HNO3, If we write 6HNO3 on LHS instead of 6H , Then we should write 6NO3- on RHS as 3Cu(NO3)2.Thus the balanced equation becomes,3Cu 8HNO3 ---> 3Cu(NO3)2 2NO 4H2O
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