Balancing Chemical Equations
Problems #26 - 45

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Sixteen balance redox equations by sight

Problem #26: Ca(AlO2)2(s) + HCl (aq) ---> AlCl3(aq) + CaCl2(aq) + H2O(ℓ)

Solution:

1) Balance the Al:

Ca(AlO2)2(s) + HCl (aq) ---> 2AlCl3(aq) + CaCl2(aq) + H2O(ℓ)

2) Balance the Cl:

Ca(AlO2)2(s) + 8HCl (aq) ---> 2AlCl3(aq) + CaCl2(aq) + H2O(ℓ)

3) Balance the H:

Ca(AlO2)2(s) + 8HCl (aq) ---> 2AlCl3(aq) + CaCl2(aq) + 4H2O(ℓ)

This last step takes care of the oxygen and the Ca was balanced from the start and never needed to be looked at during the above balancing.


Problem #27: C6H6 + H2O2 ---> CO2 + H2O

Solution:

1) Balance the carbon:

C6H6 + H2O2 ---> 6CO2 + H2O

The problem now becomes one of balancing the H and the O at the same time. What I plan to do is pick one and balance it and see what effect it has on the other. I pick hydrogen.

2) First hydrogen attempt:

C6H6 + H2O2 ---> 6CO2 + 4H2O

oxygen on left = 2; on right = 16. Difference = 14

3) Second hydrogen attempt:

C6H6 + 2H2O2 ---> 6CO2 + 5H2O

oxygen on left = 4; on right = 17. Difference = 13

Hey, look. The difference went down by 1. That sparks an idea.

4) Third hydrogen attempt:

C6H6 + 15H2O2 ---> 6CO2 + 18H2O

oxygen on left = 30; on right = 30. Difference = 0

Are the hydrogens still balanced? Yes, 36 on each side.

We have a winner!

Can you work the same idea, but with oxygen? Of course you can. After this:

C6H6 + H2O2 ---> 6CO2 + H2O

You do this:

C6H6 + (13/2)H2O2 ---> 6CO2 + H2O

Get rid of the fraction:

2C6H6 + 13H2O2 ---> 12CO2 + 2H2O

hydrogen on left = 38; on right = 4. Difference = 34

Try new numbers that still keep the oxygen in balance:

2C6H6 + 14H2O2 ---> 12CO2 + 4H2O

hydrogen on left = 40; on right = 8. Difference = 32

The difference went down by 2

Try new numbers:

2C6H6 + 30H2O2 ---> 12CO2 + 36H2O

hydrogen on left = 72; on right = 72. Difference = 0

Are the oxygens still balanced? Yes, there are 60 on each side.

Divide through by 2:

C6H6 + 15H2O2 ---> 6CO2 + 18H2O

Problem #28: Fe + O2 + H2O ---> Fe(OH)2

Solution:

As the equation is presented, the Fe and the H are already balanced. Let's leave them alone, if at all possible.

When we examine the O, we see three on the left and two on the right. Is there any way to decrease the oxygen on the left?

The answer is yes, we will use a 1/2, as follows:

Fe + (1/2)O2 + H2O ---> Fe(OH)2

This creates a total of 2 O on the left-hand side and the equation is balanced.

The accepted way to present an equation is with only whole-number coefficients, so we multiply through by a factor of 2 to arrive at the final answer:

2Fe + O2 + 2H2O ---> 2Fe(OH)2


Problem #29: MgNH4PO4 ---> Mg2P2O7 + NH3 + H2O

Solution:

1) We can balance the Mg and the P at the same time:

2MgNH4PO4 ---> Mg2P2O7 + NH3 + H2O

Since the Mg and the P only occur in one place on each side, that's the best place to start. The N is also in one place on each side, but it started out balanced and now it's not. So, . . .

2) Balance the N:

2MgNH4PO4 ---> Mg2P2O7 + 2NH3 + H2O

3) Balance H or O next?

You can pick either one, they are in equivalent circumstances. Each is in one place on the left and two places on the right.

Examine each in turn to verify they are both balanced.

We're done!


Problem #30: AlCl3(s) + Ca3N2(s) ---> AlN(s) + CaCl2(s)

Solution #1:

1) Balance the Ca:

AlCl3(s) + Ca3N2(s) ---> AlN(s) + 3CaCl2(s)

2) Balance the N:

AlCl3(s) + Ca3N2(s) ---> 2AlN(s) + 3CaCl2(s)

3) Balance the Al:

2AlCl3(s) + Ca3N2(s) ---> 2AlN(s) + 3CaCl2(s)

Note that the Cl is also balanced by balancing the Al.

Solution #2:

1) Balance the Cl by using two coefficients:

2AlCl3(s) + Ca3N2(s) ---> AlN(s) + 3CaCl2(s)

Note that this balances the Ca.

2) The Al and the N are balanced at the same time:

2AlCl3(s) + Ca3N2(s) ---> 2AlN(s) + 3CaCl2(s)

Problem #31: HNCO ---> C3N3(NH2)3 + CO2

Solution:

1) Ignore the C and the O. Balance the H and the N (because they are both in one place on the right and one place on the left):

6HNCO ---> C3N3(NH2)3 + CO2

I agree that O is in only one place on the left and one place on the right. However, trying to balance it with 2HNCO then creates problems for the H and the N, as well as then requiring 2CO2, which then requires 4HNCO. You eventually get to the right answer that way, but it's more of a hassle. How do you know in advance to focus on the H and the N? The answer is by experience and, yes, it is why I have so many examples.

2) Balance the O:

6HNCO ---> C3N3(NH2)3 + 3CO2

This also balances the C and the equation is now balanced.


Problem #32: This scary looking equation:

K4Fe(CN)6 + H2SO4 + H2O ---> K2SO4 + FeSO4 + (NH4)2SO4 + CO

looks like it might be a complicated redox equation involving three half-reactions (based on the Fe, the C and the N). However, when you examine the oxidation state of each element, you find that nothing changes.

Solution:

1) K, Fe, N and C all go from one reactant to one product. Balance them all:

K4Fe(CN)6 + H2SO4 + H2O ---> 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO

2) Sulfate is in only one place as a reactant. Add up the sulfates on the right-hand side and balance:

K4Fe(CN)6 + 6H2SO4 + H2O ---> 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO

3) The only elements left are the H in the ammonium sulfate and the O in the carbon monoxide. Count them up to get 24 H and 6 O. 12 H are accounted for in the sulfuric acid, so balance the rest (12 H and 6 O) using the water on the left:

K4Fe(CN)6 + 6H2SO4 + 6H2O ---> 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO

Problem #33: KAlSi3O8 + CO2 + H2O ---> Al2Si2O5(OH)4 + H4SiO4 + KHCO3

Solution:

Looks pretty intimidating, doesn't it? Let's see what happens.

1) Balance the Al:

2KAlSi3O8 + CO2 + H2O ---> Al2Si2O5(OH)4 + H4SiO4 + KHCO3

2) Balance the Si:

2KAlSi3O8 + CO2 + H2O ---> Al2Si2O5(OH)4 + 4H4SiO4 + KHCO3

3) Balance the K:

2KAlSi3O8 + CO2 + H2O ---> Al2Si2O5(OH)4 + 4H4SiO4 + 2KHCO3

4) Balance the C:

2KAlSi3O8 + 2CO2 + H2O ---> Al2Si2O5(OH)4 + 4H4SiO4 + 2KHCO3

5) Balance the H:

2KAlSi3O8 + 2CO2 + 11H2O ---> Al2Si2O5(OH)4 + 4H4SiO4 + 2KHCO3

You may check the oxygens if you wish. They are balanced.

Comment: I picked the Al to start because it was in one place on the left and one place on the right AND it was out of balance. True, the Si was also out of balance, but it was in one place on the left and two places on the right. Al was the simpler one to balance, so I started there.

I went with the H at the end because it was simpler than the oxygen. I had all my coefficients on the right and H was in one place on the left. Once I balanced the H, the oxygen followed right along in being balanced.


Problem #34: CH3CCl3 + O2 ---> CO2 + H2O + Cl2

1) Note the Cl, with three on the left and 2 on the right. Do this:

2CH3CCl3 + O2 ---> CO2 + H2O + 3Cl2

2) Balance C:

2CH3CCl3 + O2 ---> 4CO2 + H2O + 3Cl2

3) Balance H:

2CH3CCl3 + O2 ---> 4CO2 + 3H2O + 3Cl2

4) Balance O:

2CH3CCl3 + (11/2)O2 ---> 4CO2 + 3H2O + 3Cl2

5) Clear fraction

4CH3CCl3 + 11O2 ---> 8CO2 + 6H2O + 6Cl2

Problem #35: CO + Fe3O4 ---> CO2 + Fe

The problem is that, in order to balance the oxygen, I must use the same coefficient in front of the CO and the CO2. Obviously one does not work. Let's try 2:

2CO + Fe3O4 ---> 2CO2 + Fe

Did not work, but notice that, when using 1, the gap is three oxygen. When there is a 2, the gap has closed to two oxygen. Let's try three:

3CO + Fe3O4 ---> 3CO2 + Fe

Still not balanced by the gap is now one oxygen. There is a pattern! Use 4:

4CO + Fe3O4 ---> 4CO2 + Fe

The oxygen is now balanced and the carbon is also balanced, so finish off the iron:

4CO + Fe3O4 ---> 4CO2 + 3Fe

Problem #36: H2S + CO -----> CH3COOH + S8

1) Balance the H:

2H2S + CO -----> CH3COOH + S8

2) Balance the S:

2H2S + CO -----> CH3COOH + (1/4)S8

3) Balance the C and the O at the same time:

2H2S + 2CO -----> CH3COOH + (1/4)S8

4) Clear the fraction:

8H2S + 8CO -----> 4CH3COOH + S8

Problem #37: H2S(g) + SO2(g) ---> S8(s) + H2O(g)

1) Balance the O (because it's on one side on the left and one side on the right):

H2S(g) + SO2(g) ---> S8(s) + 2H2O(g)

2) Balance the H (it was balanced before, but balancing the O messed it up):

2H2S(g) + SO2(g) ---> S8(s) + 2H2O(g)

3) Balance the S (the only element left unbalanced):

2H2S(g) + SO2(g) ---> (3/8)S8(s) + 2H2O(g)

4) Clear the fraction:

16H2S(g) + 8SO2(g) ---> 3S8(s) + 16H2O(g)

Fractions with a denominator of 8 sure have a tendency to show up when S8 is involved. :-)


Problem #38: FeSO4 ---> Fe2O3 + SO2 + SO3

Solution #1:

1) Try breaking it up:

Left: Fe-1, S-1, O-4
Right: Fe-2, S-2, O-8

2) So you need to multiply everything on the left by 2 to balance it:

2FeSO4 ---> Fe2O3 + SO2 + SO3

Note: the ChemTeam saw the above solution scrawled on an Internet cave wall and thought it was interesting. Hope you do too!

Solution #2:

Put a 2 in front of the FeSO4 in order to balance the iron. Then, as you examine the equation, you see that everything else is balanced.


Problem #39: Pb(NO3)2 + FeCl3 ---> Fe(NO3)3 + PbCl2

Solution:

Comment: notice that the Pb and the Fe are already balanced, so let us turn our attention elsewhere.

1) Balance the nitrates:

3Pb(NO3)2 + FeCl3 ---> 2Fe(NO3)3 + PbCl2

Notice that this put the Pb and the Fe out of balance.

2) Balance the chlorides:

3Pb(NO3)2 + 2FeCl3 ---> 2Fe(NO3)3 + 3PbCl2

And in so doing, the Pb and the Fe are both balanced.

Comment: I could have done the chlorides first:

Pb(NO3)2 + 2FeCl3 ---> Fe(NO3)3 + 3PbCl2

then the nitrates:

3Pb(NO3)2 + 2FeCl3 ---> 2Fe(NO3)3 + 3PbCl2

Problem #40: C3H5(NO3)3 ---> CO2 + H2O + N2 + O2

Solution:

1) Balance the hydrogen first:

2C3H5(NO3)3 -----> CO2 + 5H2O + N2 + O2

I picked the hydrogens to go first because I realized, if I started with the carbon, I'd have to go back and change its coefficient. How did I know this? The subscripts of 5 and 2. I know that only a 2 on the left and a 5 on the right will balance the H. I could have started with C, but then I would have to go back and change the carbon coefficients after I finished balancing the H.

2) Now balance the carbon:

2C3H5(NO3)3 -----> 6CO2 + 5H2O + N2 + O2

3) Balance the nitrogen:

2C3H5(NO3)3 -----> 6CO2 + 5H2O + 3N2 + O2

4) Eighteen O on the left, nineteen on the right, so balance the O on the right by removing one (O2 to (1/2)O2):

2C3H5(NO3)3 -----> 6CO2 + 5H2O + 3N2 + (1/2)O2

5) Multiply through by 2 for the final answer:

4C3H5(NO3)3 -----> 12CO2 + 10H2O + 6N2 + O2

Problem #41: CO(NH2)2 + NO2 ---> CO2 + H2O + N2

Solution:

The key is to leave balancing the oxygen to the end. Notice that, as written, C and N are already balanced, so we will start out with the hydrogen. Also, leave the nitrogen to the end, after the oxygen.

1) Here is my first attempt to balance the H:

CO(NH2)2 + NO2 ----> CO2 + 2H2O + N2

However, this means there really isn't any way to balance the oxygen. This is because any atttempt to balance the N (using the NO2) winds up affecting the oxygens. However, then you affect the C when you try to balance the O. It's pretty bad!

2) However, let me try this:

CO(NH2)2 + NO2 ----> CO2 + 4H2O + N2

then this to balance the H:

2CO(NH2)2 + NO2 ----> CO2 + 4H2O + N2

Why did I go from 2 to 4 on the H2O? Because I knew that my H on the left would only balance in steps of 4, so the next choice was 8 H and I did that with the 4 in front of the H2O. That meant a 2 in front of the CO(NH2)2 and, hopefully, that's clear to you.

3) Balance the C

2CO(NH2)2 + NO2 ----> 2CO2 + 4H2O + N2

4) Now look at the O on the right. I have 8. I can balance the O now:

2CO(NH2)2 + 3NO2 ----> 2CO2 + 4H2O + N2

5) Now the N:

2CO(NH2)2 + 3NO2 ----> 2CO2 + 4H2O + (7/2)N2

6) Multiply through by 2 and it's done.


Problem #42: C2H2 + O2 ---> CO2 + H2O

Solution:

Balance C:

C2H2 + O2 ---> 2CO2 + H2O

H is already balanced. Balance O with 5 oxygens on the left:

C2H2 + (5/2)O2 ---> 2CO2 + H2O

Clear fraction:

2C2H2 + 5O2 ---> 4CO2 + 2H2O

Problem #43: Ag2O + NH4OH + NH4NO3 ----> [Ag(NH3)2]NO3 + H2O

Solution:

1) Balance the silver:

Ag2O + NH4OH + NH4NO3 ----> 2[Ag(NH3)2]NO3 + H2O

2) Balance the nitrate, as a group:

Ag2O + NH4OH + 2NH4NO3 ----> 2[Ag(NH3)2]NO3 + H2O

3) Note that the source of the 4 ammonia on the right side are the ammonium ions on the left. Finish balancing the ammonia:

Ag2O + 2NH4OH + 2NH4NO3 ----> 2[Ag(NH3)2]NO3 + H2O
4) Balance the left-over hydrogens and oxygens:
Ag2O + 2NH4OH + 2NH4NO3 ----> 2[Ag(NH3)2]NO3 + 3H2O

Why a three in front of the water? Let's see what extra existed on the left side before placing the three:

a) one oxygen from the Ag2O
b) two H and two OH from 2NH4OH (this compound was the source of two ammonia, everything else still needs to be balanced
c) two H from 2NH4NO3

The total left over is 6 H and 3 O, leading to the three in front of the H2O on the right.


Problem #44: HClO4 + P4O10 ---> H3PO4 + Cl2O7

Solution:

Since oxygen is in all four compounds, ignore it until the end.

1) Balance the P:

HClO4 + P4O10 ---> 4H3PO4 + Cl2O7

2) Balance the H:

12HClO4 + P4O10 ---> 4H3PO4 + Cl2O7

3) Balance the Cl:

12HClO4 + P4O10 ---> 4H3PO4 + 6Cl2O7

4) Check the oxygen:

48 + 10 = 16 + 42

It's balanced.


Problem #45: C4H10S + O2 ---> CO2 + H2O + SO2

Solution:

1) Balance the carbon:

C4H10S + O2 ---> 4CO2 + H2O + SO2

2) Balance the hydrogen:

C4H10S + O2 ---> 4CO2 + 5H2O + SO2

3) Balance the oxygen:

C4H10S + (15/2)O2 ---> 4CO2 + 5H2O + SO2

4) Clear the fractional coefficient:

2C4H10S + 15O2 ---> 8CO2 + 10H2O + 2SO2

Notice how I completely ignored the sulfur? That's because, from the beginning, it was already balanced. As I went through the steps, it remained balanced the entire time, so I never had to deal with it.


Fifteen examples      Problems 1-10      Problems 11-25      Problems 46-65
Six "balancing by groups" problems      Only the problems      Return to Equations Menu
Sixteen balance redox equations by sight