# really tough balancing equationsWatch

#1
S + HNO3 =>H2SO4 + NO2 + H2O

I've been trying to work this out for ages, but I couldn't solve it. I'm so worried that a question like this might come up in the exam and I'll just end up getting stuck. Can someone help me out, and explain how they got to the answer?
0
5 years ago
#2
Doubt you would be asked this in an exam - at least upto AS level chem. Dno about boards other than OCr though. But yeah, that is a ***** to balance.
0
5 years ago
#3
Just take one element at a time and try to balance in each step. For example there is one S on both sides at present. The hydrogens are unbalanced, you've got one on the right and 4 on the left so:

S + 4HNO3 ---> H2SO4 + NO2 + H2O

Now the Hydrogens are balanced...just keep playing with it
5 years ago
#4
I get S + 6HNO3 --> H2SO4 + 6NO2 + 2H2O

I tend to work them out via trial and improvement. I know there has to be an even number in front of HNO3 since there is an even number of hydrogen atoms and I know I have the same amount of NO2 since it's the only product with nitrogen. Then it was just a matter of trying a couple of combinations to balance the hydrogens and oxygens.

Haha, I don't really have a system I work by, but I'd try to balance the atoms that only appear once in the reactants and products (such as S and N in this case) and work from there.

Posted from TSR Mobile
0
5 years ago
#5
Hee, I love balancing equations. My methods might not help, but on the offchance:

1. I figured out there must be an even number of HNO3 molecules on the left because no matter what multiples I put on the right, I need an even number of H on the right.
2. I found the base amount of oxygen on the right is 7 atoms, but to transfer it back to the left, I'd need multiples of 3. Since 9 O would give me 3 H, an odd number, I decided to make it 12 O, giving me 4 HNO3 on the left.
3. If I have 4 HNO3, I'll need 4 NO2 on the right to balance the N.
4. I counted up the Oxygens on the right - H2SO4 + 4 NO2 = 12 oxygens, the same as I have on the left. But... the water needs to come from somewhere, so back to the left we go.
5. I added another two HNO3 to the left to keep the H even and then adjusted the water to balance the extra H and O.

Of course if it comes up in the exam, skip it or make a deliberate mistake - if a whole question hinges on it you may get an Error Carried Forward mark out of it
0
#6
(Original post by EierVonSatan)
Just take one element at a time and try to balance in each step. For example there is one S on both sides at present. The hydrogens are unbalanced, you've got one on the right and 4 on the left so:

S + 4HNO3 ---> H2SO4 + NO2 + H2O

Now the Hydrogens are balanced...just keep playing with it
I did, but I got no where , it's like an endless cycle, when I change one I need to change another, then another, then another.

But I guess it can't hurt to have one more try.
0
#7
(Original post by AbiC)
I get S + 6HNO3 --> H2SO4 + 6NO2 + 2H2O

I tend to work them out via trial and improvement. I know there has to be an even number in front of HNO3 since there is an even number of hydrogen atoms and I know I have the same amount of NO2 since it's the only product with nitrogen. Then it was just a matter of trying a couple of combinations to balance the hydrogens and oxygens.

Haha, I don't really have a system I work by, but I'd try to balance the atoms that only appear once in the reactants and products (such as S and N in this case) and work from there.

Posted from TSR Mobile

(Original post by Qaisha)
Hee, I love balancing equations. My methods might not help, but on the offchance:

1. I figured out there must be an even number of HNO3 molecules on the left because no matter what multiples I put on the right, I need an even number of H on the right.
2. I found the base amount of oxygen on the right is 7 atoms, but to transfer it back to the left, I'd need multiples of 3. Since 9 O would give me 3 H, an odd number, I decided to make it 12 O, giving me 4 HNO3 on the left.
3. If I have 4 HNO3, I'll need 4 NO2 on the right to balance the N.
4. I counted up the Oxygens on the right - H2SO4 + 4 NO2 = 12 oxygens, the same as I have on the left. But... the water needs to come from somewhere, so back to the left we go.
5. I added another two HNO3 to the left to keep the H even and then adjusted the water to balance the extra H and O.

Of course if it comes up in the exam, skip it or make a deliberate mistake - if a whole question hinges on it you may get an Error Carried Forward mark out of it
0
5 years ago
#8
(Original post by tammie123)
S + HNO3 =>H2SO4 + NO2 + H2O

I've been trying to work this out for ages, but I couldn't solve it. I'm so worried that a question like this might come up in the exam and I'll just end up getting stuck. Can someone help me out, and explain how they got to the answer?
You can solve this algebraically. Assume a value for one of the balancing numbers and then give the others a letter. i.e:

aS + bHNO3 =>cH2SO4 + dNO2 + H2O

You know that a = c to balance the sulfurs and that b=d to balance the nitrogens:

aS + bHNO3 =>aH2SO4 + bNO2 + H2O

Now equate the number of hydrogens on both sides:

b = 2a + 2

And for Oxygens

3b = 4a + 2b + 1
=> b = 4a + 1

Solve the two equations for a and b,

2a + 2 = 4a + 1
=> a = 1/2, b = 3.

Therefore,

1/2S + 3HNO3 => 1/2H2SO4 + 3NO2 + H2O

Multiply everything by 2 for neatness,

S + 6HNO3 => H2SO4 + 6NO2 + 2H2O
5
#9
(Original post by Upper Echelon)
You can solve this algebraically. Assume a value for one of the balancing numbers and then give the others a letter. i.e:

aS + bHNO3 =>cH2SO4 + dNO2 + H2O

You know that a = c to balance the sulfurs and that b=d to balance the nitrogens:

aS + bHNO3 =>aH2SO4 + bNO2 + H2O

Now equate the number of hydrogens on both sides:

b = 2a + 2

And for Oxygens

3b = 4a + 2b + 1
=> b = 4a + 1

Solve the two equations for a and b,

2a + 2 = 4a + 1
=> a = 1/2, b = 3.

Therefore,

1/2S + 3HNO3 => 1/2H2SO4 + 3NO2 + H2O

Multiply everything by 2 for neatness,

S + 6HNO3 => H2SO4 + 6NO2 + 2H2O
This methods awesome! Did you come up with this yourself? It's so much easier and more reliable than the trial and error method.
0
5 years ago
#10
(Original post by tammie123)
This methods awesome! Did you come up with this yourself? It's so much easier and more reliable than the trial and error method.
The more usual method is to use the redox half-equations.

S + HNO3 => H2SO4 + NO2 + H2O

Sulphur gets oxidised from zero oxidation state to +6 hence it must have lost 6 electrons. Use water and hydrogen ions to balance the other atoms:

S + 4H2O --> H2SO4 + 6e + 6H+

nitrogen changes from oxidation state +5 to +4 hence it must have gained 1 electron. Once again use water and hydrogen ions for the other atoms:

HNO3 + 1e + H+ --> NO2 + H2O

Now equalise the electrons by multiplying the second half-equation by 6. Then add them together.

S + 4H2O --> H2SO4 + 6e + 6H+
6HNO3 + 6e + 6H+ --> 6NO2 + 6H2O
S + 4H2O + 6HNO3 + 6e + 6H+ --> H2SO4 + 6e + 6H+ + 6NO2 + 6H2O

finally cancel out common terms on either side

S + 6HNO3 --> H2SO4 + 6NO2 + 2H2O
1
#11
(Original post by charco)
The more usual method is to use the redox half-equations.

S + HNO3 => H2SO4 + NO2 + H2O

Sulphur gets oxidised from zero oxidation state to +6 hence it must have lost 6 electrons. Use water and hydrogen ions to balance the other atoms:

S + 4H2O --> H2SO4 + 6e + 6H+

nitrogen changes from oxidation state +5 to +4 hence it must have gained 1 electron. Once again use water and hydrogen ions for the other atoms:

HNO3 + 1e + H+ --> NO2 + H2O

Now equalise the electrons by multiplying the second half-equation by 6. Then add them together.

S + 4H2O --> H2SO4 + 6e + 6H+
6HNO3 + 6e + 6H+ --> 6NO2 + 6H2O
S + 4H2O + 6HNO3 + 6e + 6H+ --> H2SO4 + 6e + 6H+ + 6NO2 + 6H2O

finally cancel out common terms on either side

S + 6HNO3 --> H2SO4 + 6NO2 + 2H2O
Thank you I think I'll stick to this method, it's the one I'm most familiar with as we covered it in class (,so I'm less likely to get it wrong in the exam ).
0
10 months ago
#12
Why did you not label your H2O at the end with an e but you labeled the rest
0
10 months ago
#13
(Original post by charco)
The more usual method is to use the redox half-equations.

S + HNO3 => H2SO4 + NO2 + H2O

Sulphur gets oxidised from zero oxidation state to +6 hence it must have lost 6 electrons. Use water and hydrogen ions to balance the other atoms:

S + 4H2O --> H2SO4 + 6e + 6H+

nitrogen changes from oxidation state +5 to +4 hence it must have gained 1 electron. Once again use water and hydrogen ions for the other atoms:

HNO3 + 1e + H+ --> NO2 + H2O

Now equalise the electrons by multiplying the second half-equation by 6. Then add them together.

S + 4H2O --> H2SO4 + 6e + 6H+
6HNO3 + 6e + 6H+ --> 6NO2 + 6H2O
S + 4H2O + 6HNO3 + 6e + 6H+ --> H2SO4 + 6e + 6H+ + 6NO2 + 6H2O

finally cancel out common terms on either side

S + 6HNO3 --> H2SO4 + 6NO2 + 2H2O
Why did you not label your H2O at the end with an E
0
10 months ago
#14
(Original post by Coldcash90900)
Why did you not label your H2O at the end with an E
???
1
10 months ago
#15
(Original post by Upper Echelon)
You can solve this algebraically. Assume a value for one of the balancing numbers and then give the others a letter. i.e:

aS + bHNO3 =>cH2SO4 + dNO2 + H2O

You know that a = c to balance the sulfurs and that b=d to balance the nitrogens:

aS + bHNO3 =>aH2SO4 + bNO2 + H2O

Now equate the number of hydrogens on both sides:

b = 2a + 2

And for Oxygens

3b = 4a + 2b + 1
=> b = 4a + 1

Solve the two equations for a and b,

2a + 2 = 4a + 1
=> a = 1/2, b = 3.

Therefore,

1/2S + 3HNO3 => 1/2H2SO4 + 3NO2 + H2O

Multiply everything by 2 for neatness,

S + 6HNO3 => H2SO4 + 6NO2 + 2H2O
Why didn't you label the h2O you had at the end with an e
0
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