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If you're just starting to learn how to balance chemical equations, I would like to suggest doing some easier ones first. Then, come back here. If not, then welcome. Teachers do like to teach the easier ones then ask ones like 2, 3, 4, 7, 10 and 13.

Problem #1: 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 #2: Al + KOH + H₂SO₄ + H₂O ----> KAl(SO₄)₂ · 12H₂O + H₂

This problem can be solved by eliminating the water from both sides and then adding it back in when the equation is balanced. I do that here. There is a another problem involving KAl(SO₄)₂ · 12H₂O in that file.

Solution:

1) Balance the water of hydration:

Al + KOH + H₂SO₄ + 12H₂O ----> KAl(SO₄)₂ · 12H₂O + H₂

2) Balance the sulfates:

Al + KOH + 2H₂SO₄ + 12H₂O ----> KAl(SO₄)₂ · 12H₂O + H₂

3) Notice that I reduce the water on the left by one:

Al + KOH + 2H₂SO₄ + 11 H₂O ----> KAl(SO₄)₂ · 12H₂O + H₂

I'm doing this because I still need to take care of the hydroxide and the way to do that is by adding a hydrogen to it and make water. This water will replace the one water that I eliminated from the left-hand side.

4) Balance the hydrogens:

Al + KOH + 2H₂SO₄ + 11 H₂O ----> KAl(SO₄)₂ · 12H₂O + (3/2) H₂

There were 4 hydrogens in the sulfuric acid that had to be balanced. One of the H reacts with the hydroxide to replace the twelfth water I eliminated on the left-hand side above. The 3/2 on the right balances the other three that are left over.

5) Multiply through by two to clear the fractional coefficient:

2Al + 2KOH + 4H₂SO₄ + 22 H₂O ----> 2KAl(SO₄)₂ · 12H₂O + 3 H₂

Problem #3: 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 #4: 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 #5: NH₄NO₃ (s) ---> N₂ (g) + O₂ (g) + H₂O (g)

Solution:

1) Balance the hydrogen:

NH₄NO₃ (s) ---> N₂ (g) + O₂ (g) + 2H₂O (g)

2) Too many O on the right, so reduce the amount:

NH₄NO₃ (s) ---> N₂ (g) + (1/2)O₂ (g) + 2H₂O (g)

3) Multiply by through by 2:

2NH₄NO₃ (s) ---> 2N₂ (g) + O₂ (g) + 4H₂O (g)

Problem #6: Fe₂O₃(s) + C(s) ---> Fe(s) + CO₂(g)

Solution:

1) Balance the iron:

Fe₂O₃(s) + C(s) ---> 2Fe(s) + CO₂(g)

2) Balance the oxygen:

Fe₂O₃(s) + C(s) ---> 2Fe(s) + (3/2)CO₂(g)

3) Balance the carbon:

Fe₂O₃(s) + (3/2)C(s) ---> 2Fe(s) + (3/2)CO₂(g)

Note the three-halves in front of the C and the CO₂. What's that you say? You can't have a three-halves of an atom? Ah, just you wait.

4) Multiply through by two for the final answer:

2Fe₂O₃(s) + 3C(s) ---> 4Fe(s) + 3CO₂(g)

Comment: using the three-halves was just a mathematical artifice to balance the equation. The chemical reality of atoms reacting in ratios of small whole numbers is reflected in the final answer.

Problem #7: C₅H₁₁NH₂ + O₂ ---> CO₂ + H₂O + NO₂

Solution:

1) Balance the hydrogens first:

2C₅H₁₁NH₂ + O₂ ---> CO₂ + 13H₂O + NO₂

Notice that I used a 2 in front of C₅H₁₁NH₂. That's because I knew that there are 13 hydrogens in the C₅H₁₁NH₂ and that meant a 13/2 in front of the water. I knew I'd have to eventually clear the 13/2, so I decided to do so right at the start.

2) Balance the nitrogen and the carbon:

2C₅H₁₁NH₂ + O₂ ---> 10CO₂ + 13H₂O + 2NO₂

3) Oxygen:

2C₅H₁₁NH₂ + (37/2)O₂ ---> 10CO₂ + 13H₂O + 2NO₂

4) Multiply through by 2 for:

4, 37 ---> 20, 26, 4

Problem #8: CO₂ + S₈ ---> CS₂ + SO₂

Solution:

1) The only thing not balanced already is the S:

CO₂ + (3/8)S₈ ---> CS₂ + SO₂

Most of the time the fraction used to balance is one with a 2 in the denominator: 1/2 or 5/2 or 13/2. Not too often does one see 3/8. Pretty tricky!

2) Multiply through by 8:

8CO₂ + 3S₈ ---> 8CS₂ + 8SO₂

Problem #9: NO₂ + H₂O ---> HNO₃ + NO

Solution:

1) The H is in one compound on the left and one compound on the right. Balance it first:

NO₂ + H₂O ---> 2HNO₃ + NO

2) Balance the N:

3NO₂ + H₂O ---> 2HNO₃ + NO

3) As a result of balacing the nitrogen, the oxygens also are balanced. We are done!

Problem #10: 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

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 #11: Cu + HNO₃ ---> Cu(NO₃)₂ + NO₂ + H₂O

Solution:

1) The key is to see that the HNO₃ coefficient must be twice the value of the H₂O coefficient. This is because H only appears on the left in the HNO₃ and appears in the right only in the H₂O. Like this:

Cu + 2HNO₃ ---> Cu(NO₃)₂ + NO₂ + H₂O

2) The above will not work (even though the H is now balanced). Even though the N is balanced, there is no way to balance the O (since there is already a 2 in front of the nitric acid), so do this:

Cu + 4HNO₃ ---> Cu(NO₃)₂ + NO₂ + 2H₂O

What happened here is that the 2HNO₃ & H₂O combination will not work, so I went to a combination of 4HNO₃ & 2H₂O. I will continue balancing, hoping the the 4 & 2 coefficients will work.

3) Now, balance the N:

Cu + 4HNO₃ ---> Cu(NO₃)₂ + 2NO₂ + 2H₂O

4) Check the oxygen:

twelve on each side. It's balanced.

Problem #12: 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.

Problem #13: NaBH₄ + BF₃ ---> NaBF₄ + B₂H₆

Solution:

1) Balance the F:

NaBH₄ + 4BF₃ ---> 3NaBF₄ + B₂H₆

2) Balance the H:

3NaBH₄ + 4BF₃ ---> 3NaBF₄ + 2B₂H₆

Notice that the Na on the left and the Na on the right are also balanced. Count up the B to make sure it is balanced also.

Why did I do what I did? Note that the F only occurs in one compound on the left and one compound on the right, so I ignored the boron and concentrated on the fluorine.

Then, I did the same thing for the H. It's in only one compound on the left as well as only one compound on the right. So, once again, I ignored everything else and balanced the H. Then, I checked the Na and the B and determined that the equation was now balanced.

Problem #14: C₅H₇O + O₂ ---> CO₂ + H₂O

Solution:

1) Balance carbon:

C₅H₇O + O₂ ---> 5CO₂ + H₂O

2) Balance hydrogen:

2C₅H₇O + O₂ ---> 10CO₂ + H₂O

then

2C₅H₇O + O₂ ---> 10CO₂ + 7H₂O

3) Balance oxygen:

2C₅H₇O + (25/2)O₂ ---> 10CO₂ + 7H₂O

4) clear fraction:

4C₅H₇O + 25O₂ ---> 20CO₂ + 14H₂O

Balancing the hydrogens in step two was interesting due to the seven H present on the left-hand side. The problem is that the hydrogens only come in even numbers of the right-hand side, due to the H₂O. So, the solution is to make an even number of hydrogens on the left-hand side with the coefficient of two in front of the C₅H₇O.

Problem #15: NaOH + P₄ + H₂O ---> NaH₂PO₂ + PH₃

Solution:

1) Phosphorus:

NaOH + P₄ + H₂O ---> 3NaH₂PO₂ + PH₃

I picked using the P in the NaH₂PO₂ rather than the PH₃ because I knew I'd have to do both H and O. Putting a 3 in front of the NaH₂PO₂ gives me more O to work with.

2) Sodium:

3NaOH + P₄ + H₂O ---> 3NaH₂PO₂ + PH₃

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

3NaOH + P₄ + 3H₂O ---> 3NaH₂PO₂ + PH₃

Problem #16: NBr₃ + NaOH ----> N₂ + NaBr +HOBr

Solution:

1) Nitrogen:

2NBr₃ + NaOH ----> N₂ + NaBr + HOBr

Some explanation: the next step is to balance the Br, however there are issues. Notice how the Na and the OH and both in one compound on the left, but are in TWO compounds on the right. The consequence of that is that the coefficient in front of the NaBr must be the same as the coefficient in front of the HOBr. At the same time, those two coefficients MUST provide six bromines.

2) That being said, here's the next step:

2NBr₃ + NaOH ----> N₂ + 3NaBr + 3HOBr

And that's the answer.

Problem #17: VO + Fe₂O₃ ---> FeO + V₂O₅

Solution:

The oxygen issue is complex, it being in every compound in the problem. So, let's do everything else and see what happens.

1) Balance the V:

2VO + Fe₂O₃ ---> FeO + V₂O₅

2) Balance the Fe:

2VO + Fe₂O₃ ---> 2FeO + V₂O₅

This does not balance the oxygens.

3) Let us not give up hope. Try a new set of coefficients in front of the Fe:

2VO + 2Fe₂O₃ ---> 4FeO + V₂O₅

This does not balance the oxygens. However, something has happened. The gap in the oxygens has closed. With Fe coefficients of 1 & 2, there was a two oxygen gap. With Fe coeccifients of 2 & 4, there is a one oxygen gap. Let us continue.

4) Try a new set of coefficients in front of the Fe:

2VO + 3Fe₂O₃ ---> 6FeO + V₂O₅

It's balanced.

Problem #18: Fe₂O₃(s) + CO(g) ---> Fe(s) + CO₂(g)

Solution:

1) Balance the Fe:

Fe₂O₃(s) + CO(g) ---> 2Fe(s) + CO₂(g)

To balance the oxygen, we follow the same path as balancing the oxygens in the problem just above. Notice that the carbon is balanced. What we will do is try new coefficients that keep the carbon balanced.

2) Trial #1:

Fe₂O₃(s) + 2CO(g) ---> 2Fe(s) + 2CO₂(g)

This does not balance the oxygens, but something good does happen. When the carbon coefficients were 1 & 1, there was a two oxygen gap. With carbon coefficients of 2 & 2, the gap is now only one oxygen.

3) Trial #2:

Fe₂O₃(s) + 3CO(g) ---> 2Fe(s) + 3CO₂(g)

It's balanced.

Problem #19: S₂H₅ + O₂ ---> SO₂ + H₂O

Solution:

1) Balance the H:

2S₂H₅ + O₂ ---> SO₂ + 5H₂O

Notice how I did it with two coefficients at the same time. That is because 10 is the least common multiple between 2 and 5.

2) Balance the S:

2S₂H₅ + O₂ ---> 4SO₂ + 5H₂O

3) Balance the O:

2S₂H₅ + (13/2)O₂ ---> 4SO₂ + 5H₂O

4) Clear the fraction:

4S₂H₅ + 13O₂ ---> 8SO₂ + 10H₂O

Problem #20: KOH + F₂ ---> KF + F₂O + H₂O

Solution:

1) Balance the H:

2KOH + F₂ ---> KF + F₂O + H₂O

This also balances the oxygen.

2) Balance the K:

2KOH + F₂ ---> 2KF + F₂O + H₂O

3) Balance the F:

2KOH + 2F₂ ---> 2KF + F₂O + H₂O

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