Balancing Chemical Equations
Problems #26 - 45

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

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


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


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 + 132H2O2 ---> 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


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 12, as follows:

Fe + 12O2 + 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


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


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: CdS + O2 ---> Cd + SO2


Nothing need be done. This equation is already balanced. When referring to an equation that is already balanced and nothing needs to be done, the term used is "balanced as written.'

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


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 a subscript of 3 on the left and a subscript of 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 + 112O2 ---> 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 + 14S8

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

2H2S + 2CO -----> CH3COOH + 14S8

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) ---> 38S8(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


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 (by the ChemTeam):

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 12O2):

2C3H5(NO3)3 -----> 6CO2 + 5H2O + 3N2 + 12O2

5) Multiply through by 2 for the final answer:

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

6) Notice that the two methods below each use carbon as their starting point, resulting in a multiply-through of 4 (note that method #3 multiplies by two twice). Having 14O2 is not very common and the way I chose to balance the equation did not require a 14O2.

Solution #2 (not by the ChemTeam):

Start with the least common elements. Notice that O shows up in 4 compounds, but C, H, and N only show up in 2 compounds each.

1) Starting with C:

C3H5N3O9 ---> 3CO2 + N2 + H2O + O2

2) Now H:

C3H5N3O9 ---> 3CO2 + N2 + 52H2O + O2

3) Next balance N:

C3H5N3O9 ---> 3CO2 + 32N2 + 52H2O + O2

4) Finally, balance O:

C3H5N3O9 ---> 3CO2 + 32N2 + 52H2O + 14O2

The rationale for the 14O2 starts with the fact that there are 9 oxygens on the left (but think of it as 182). On the right are 6 oxygens from the carbon dioxide (think of it as 122). Add 52 from the water to get 172. The O2 contributes 22 for a total of 192. This is 8.5 oxygen atoms, so we are short 0.5 oxygen atom on the right.

14O2 represents one-half of an O atom (remember that 12O2 represents one O atom). By the way, saying one-half (or one-quarter) of an atom is just a rhetorical device used on the way to balancing the equation with whole numbers. The fractions do not literally mean one-half (or one-quarter) of an atom.

5) Quadruple to get rid of the fractions:

4C3H5N3O9 ---> 12CO2 + 6N2 + 10H2O + O2

Solution #3 (not by the ChemTeam):

Note the use of decimals rather than fractions. This is perfectly OK, it's just a stylistic thing.

1) Balance the carbon:

C3H5N3O9 ---> 3CO2 + N2 + H2O + O2

2) Balance the nitrogen:

C3H5N3O9 ---> 3CO2 + 1.5N2 + H2O + O2

3) Balance the hydrogen:

C3H5N3O9 ---> 3CO2 + 1.5N2 + 2.5H2O + O2

4) Multiply everything by 2 to get rid of the fractions:

2C3H5N3O9 ---> 6CO2 + 3N2 + 5H2O + 2O2

5) Now you've got 2 x 9 = 18 oxygens on the left, and 6 x 2 + 5 + 2 x 2 = 21 oxygens on the right.

6) You can't change the left side or CO2 or H2O without messing up the balances for C or H, so you must adjust the O2 on the right, which has 21 − 18 = 3 too many oxygen atoms, so subtract 3 atoms of O on the right:

2C3H5N3O9 ---> 6CO2 + 3N2 + 5H2O + 0.5O2

7) Now the equation is completely balanced, but not completely whole number coefficients. To remove the fraction, multiply everything by 2 again:

4C3H5N3O9 ---> 12CO2 + 6N2 + 10H2O + O2

Solution #4 (not by the ChemTeam):

More discussion than in the above three solutions.

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


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 + 72N2

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

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


1) Balance C:

C2H2 + O2 ---> 2CO2 + H2O

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

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

3) Clear fraction:

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

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


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 #1:

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.

Solution #2:

1) I'll do both the chlorine and the phosphorous at the same time:

2HClO4 + P4O10 ---> 4H3PO4 + Cl2O7

2) Balance the hydrogen, not the oxygen:

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

3) This throws the chlorine out of balance, so rebalance it:

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

4) 58 oxygens on each side means the equation is balanced.

5) Here's an interesting wrong direction:

If you balance the Cl after the P, you get this:

2HClO4 + P4O10 ---> 4H3PO4 + Cl2O7

However, you still have to balance the H and you can only get H balanced by changing the HClO4 coefficient. In addition, by putting the 2 in front of the HClO4, you block yourself from balancing the oxygen.

So, if you go this direction, you have to see that it is a dead end and erase the 2 and go with 12.

6) When I first saw this wrong direction, I thought to add water to balance everything:

2HClO4 + P4O10 + 5H2O ---> 4H3PO4 + Cl2O7

The oxygens do balance, but now the equation has been changed from what was asked. Not allowed (unless there is a mistake).

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


1) Balance the carbon:

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

2) Balance the hydrogen:

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

3) Balance the oxygen:

C4H10S + 152O2 ---> 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.

Bonus Problem: CaF2 3Ca3(PO4)2 + H2SO4 + H2O ---> Ca(H2PO4)2 + HF + CaSO4 2H2O


1) Balance the phosphorous:

CaF2 3Ca3(PO4)2 + H2SO4 + H2O ---> 3Ca(H2PO4)2 + HF + CaSO4 2H2O

2) Balance the calcium:

CaF2 3Ca3(PO4)2 + H2SO4 + H2O ---> 3Ca(H2PO4)2 + HF + 7CaSO4 2H2O

3) Balance the sulfur:

CaF2 3Ca3(PO4)2 + 7H2SO4 + H2O ---> 3Ca(H2PO4)2 + 2HF + 7CaSO4 2H2O

I also balanced the fluorine.

4) Look at the hydrogens NOT in water and find 14 of them on each side. The only thing not balanced are the 14 waters in 7CaSO4 2H2O. Balance them and the equation is done:

CaF2 3Ca3(PO4)2 + 7H2SO4 + 14H2O ---> 3Ca(H2PO4)2 + 2HF + 7CaSO4 2H2O

Twenty examples      Probs 1-10      Probs 11-25      Probs 46-65
"Balancing by groups" problems      Only the problems      Equations Menu
Balance redox equations by sight