Balancing Equations
This tutorial will just take you through several
balancing equation problems, starting with
some pretty easy ones, and then getting
progressively more and more difficult.
Balancing Equations
1. __ H2 + __ F2  __
2 HF
H H
F
F
Starting with the H’s, we have two on the left.
and only one on the right.
So we need to put a “2” on the HF.
That gives us two H’s on both sides.
Now the F’s: We have two on the left,
And we have two on the right, because the “2” we
put in front of the HF pertains to the entire HF
molecule – two H’s and 2 F’s.
So everything is balanced.
Balancing Equations
2. __ N2 + __
3 F2  __
2 NF3
three groups
ofF 2 = two groups of 3
F
N
N
F
F
F
F
Starting with the N’s, we have two on the left.
and only one on the right.
So we need to put a “2” on the NF3.
That gives us two N’s on both sides.
Now the F’s: We have two on the left,
And how many F’s on the right? 6 If you were thinking 3, you
forgot about the 2 we put in
So we need to put a 3 on the F2. front of the NF3. 2 x 3 = 6.
That gives us six F’s on both sides.
Now everything is balanced.
Balancing Equations
3 Na2S + __ FeBr3  __
3. __
62 NaBr + __ Fe2S3
three groups of 2 = six groups of 1
Starting with the Na’s, we have two on the left.
and only one on the right.
So we need to put a “2” on the NaBr.
That gives us two Na’s on both sides.
Now the S’s: We have one on the left,
And three on the right?
So we need to put a 3 on the Na2S.
That gives us three S’s on both sides,
6
But it messes up the Na’s: now we have six Na’s on the left.
So we need to change the 2 NaBr to 6 NaBr.
Since you often have to change numbers you’ve written down, it’s always a good
idea to use a pencil or erasable pen when balancing equations!
Balancing Equations
3 Na2S + __
3. __
2 FeBr3  __
6 NaBr + __ Fe2S3
So now the Na’s are balanced again with six on each side.
Sorry, I’m not going
Now the Fe’s: we have one Fe on the left
to show you that
and two Fe’s on the right.
little rearrangement
So what should we do? Put a 2 on the FeBr3. of the atoms this
time… Hopefully
Now we have 2 Fe’s on both sides.
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And how many Br’s are on the right? Also 6 coefficients above!
So it seems everything is balanced, but with all the back
and forth and changes made, it’s best to check our answer.
Go ahead and “take inventory” on both sides.
You should have counted six Na’s, three S’s, two Fe’s and six Br’s on
each side… so it is balanced!
Balancing Equations
So now let’s practice a few relatively easy equation balancing
problems and see how you do. Try these on your own, one at a
time, and then advance the slide to see if you are right. If you get it
wrong, try to see where you made your mistake.
2 H2 + __ O2  __
2 H2O
4. __
6 K + __ N2  __
2 K3N
5. __
4 Fe + __
3 O2  __
2 Fe2O3
6. __
2 AlBr3 + __
3 I2  __
2 AlI3 + __
3 Br2
7. __
4 PF3  __ P4 + __
6 F2
8. __
4 SnF4 + __ S8  __
4 SnS2 + __
8 F2
9. __
Balancing Equations
10. __ K3BO3 + __ Sn(C2H3O2)4  __ KC2H3O2 + __ Sn3(BO3)4
Right away you are probably looking at this one and
thinking that it’s going to be really difficult. But there is a
“trick of the trade” we will use to make it much easier.
Do you notice that there are polyatomic ions that appear
on both sides?
Rather than breaking them up into their atoms, we are
going to keep them together and balance them as bundles.
For example, rather than breaking up the first one into B’s
and O’s, we will simply ask: “How many BO3 bundles are on
the left side, and how many are on the right side?”
Answers: one on the left and four on the right.
(Keep in mind, the “3” is part of the BO3 bundle.)
Balancing Equations
3 KC2H3O2 + __ Sn3(BO3)4
4 K3BO3 + __
3 Sn(C2H3O2)4  12
10. __
__
So, let’s start with the K’s. How would you balance them?
Hopefully you saw there were three on the left and only one
on the right and so you put a 3 on the KC2H3O2.
Now, as we said before, we have one BO3 on the left and
four on the right. What should we do to balance the BO3’s?
Putting a 4 on the K3BO3 makes the BO3‘s balanced but…
Now we have twelve K’s on the left. What should we do?
Were you thinking of changing the 3 on KC2H3O2 to a 12?
Good, so now how do we balance the Sn’s?
It will require a 3 on the Sn(C2H3O2)2 to balance them.
Finally, how many C2H3O2 bundles do we have on the left?
If you were thinking four, check again. It’s twelve. And with
twelve C2H3O2’s on the right as well, the equation’s balanced.
Balancing Equations
So now let’s practice a few equation balancing problems involving
these polyatomic ions.
Remember trick of the trade #1: Keep polyatomic ions together and
balance them as bundles.
11. __
2 (NH4)2SO4 + __ Pb(NO3)4  __ Pb(SO4)2 + __
4 NH4NO3
12. __ Al2(CO3)3 + __
6 HClO4  __
2 Al(ClO4)3 + __
3 H2CO3
13. __
3 Ca(NO3)2 + __
2 K3PO4  __ Ca3(PO4)2 + __
6 KNO3
14. __
4 (NH4)3BO3 + __
3 Sn(BrO3)4  __ Sn3(PO4)4 + __
12 NH4BrO3
Balancing Equations
15. __ Na + __ H2O  __ NaOH + __ H2
Sometimes when H2O is in a simple equation like the one
above, it may prove difficult to balance.
Trick of the trade #2: If you find yourself having trouble
balancing an equation that has H2O in it, try treating the
H2O like H(OH) instead, and then balance the H and the OH
completely separately: you may find it easier to balance.
Balancing Equations
H(OH)
2 H2O  __
2 Na + __
2 NaOH + __ H2
15. __
So, in the equation above, we will try this:
We can see we have one Na on each side, so they’re good.
How many H’s do we have on the left? Just one.
If you were thinking two, then you were forgetting, we are
balancing the H’s and the OH’s separately.
And how many H’s do we have on the right? Two.
Remember: don’t count the H in Na(OH).
So let’s put a 2 on the H(OH).
Now we have two OH’s on the left & only one on the right.
So we must put a 2 on the NaOH.
So are we all done? Not quite.
Now the Na’s need to be balanced again! Now we’re done!
Balancing Equations
So now let’s practice a few problems like that last one.
Remember trick of the trade #2: Sometimes it’s useful to treat H2O
as H(OH) and balance the H and the OH separately.
16. __
2 H2O + __ Ba  __ H2 + __ Ba(OH)2
17. __
2 Cr + __
6 H2O  __
2 Cr(OH)3 + __
3 H2
18. __
3 H2SO4 + __
2 Fe(OH)3  __
6 H2O + __ Fe2(SO4)3
Balancing Equations
19. __ O2 + __ C3H8  __ CO2 + __ H2O
This problem may look pretty easy, but there is one slight
speed bump you need to be aware of. Whereas in all the
problems we have tried so far, each element (or polyatomic
ion bundle) has appeared in just two places – one on each
side of the equation– here we have an element that occurs
in three places: O
Trick of the trade #3: When an element appears in more
than two places in an equation, always save it for last.
Balancing Equations
3 CO2 + __
4 H2O
5 O2 + __ C3H8  __
19. __
So rather than begin with the O’s, we will skip them and start
with the C’s. How would you balance the C’s?
Hopefully you saw that there were three C’s on the left and
only one on the right, so you put a 3 on the CO2.
Now let’s balance the H’s… Any ideas?
If you saw that there were eight H’s on the left and only two
on the right, you probably knew you needed a 4 on the H2O.
Now the O’s. You can see that there are two on the left side,
but how many O’s are there all together on the right?
Your answer should have been ten. six plus four
So with ten O’s on the right and only two on the left, you
should know what to do:
Balancing Equations
So now let’s practice a few equation balancing problems involving
elements appearing more than twice.
Remember trick of the trade #3: If an element appears more than
twice in the equation, save balancing that element for last.
20. __ C5H8 + __
7 O2  __
5 CO2 + __
4 H2O
21. __ C3H8 + 10
__ F2  __
3 CF4 + __
8 HF
22. __
2 N3H5 + __
4 O2  __
3 N2O + __
5 H2O
23. __ C9H18O + 13
__ O2  __
9 CO2 + __
9 H2O
Balancing Equations
24. __ C4H10 + __ O2  __ CO2 + __ H2O
This problem looks pretty much like the last ones you just
did, with one slight difference. The easiest way to balance
it requires a fraction. Yet balanced equations are most
often written with whole numbers only.
Trick of the trade #4: If you need to use a fraction to
balance an equation, do it, but then multiply everything by
the denominator (usually 2) to get rid of the fraction.
Balancing Equations
/2 O2  __
2 C4H10 +1313
10
48 CO2 + __
5 H2O
24. __
__
First we will balance the C’s.
Next let’s balance the H’s.
There are two O’s on the left; how many O’s on the right?
Hopefully you counted thirteen. eight plus five
And the only place we can put a number that won’t throw
everything else off is in front of the O2.
But how do you change a 2 into a 13?
Answer: by multiplying it by 6.5 …or 61/2 …or 13/2.
But because we can’t leave fractions in a balanced
equation, we will now multiply every coefficient by the
fraction’s denominator: 2.
And now the equation is balanced.
Balancing Equations
So now let’s practice a few final balancing problems involving
fractions.
Remember trick of the trade #4: Use fractions if you need to, but
then multiply through to get rid of them.
25. __
2 C2H6 + __
7 O2  __
4 CO2 + __
6 H2O
26. __
__ O2  10
__ CO2 + 10
__ H2O
2 C5H10 + 15
27. __
__ O2  __
2 N4H6 + 11
8 NO2 + __
6 H2O
28. __
__ FeP
4 Fe3P2 + __
3 P4  12
2 C3H8O + __
9 O2  __
6 CO2 + __
8 H2O
29. __
3 C6H14O2 + __
17 Fe2O3  __
18 CO2 + __
21 H2O + 34
30. __
__ Fe
These last
three are
pretty tricky,
but try your
best!