BALANCING HARD TO BALANCE EQUATIONS
COMBUSTION REACTIONS:
Many combustion reactions are hard to balance because of the oxygen in both the CO2 and H2O. Often a mixed
fraction of ½ for the O2 will balance the equation. Only thing is you can’t have a fraction for a coefficient.
Therefore, you double ALL the coefficients when this happens.
Example: Balance C4H10 + O2 → CO2 + H2O

After doing the C and H you end up with the following.
C4H10 + O2 → 4CO2 + 5H2O

Since there are 13 oxygens on the product side, the coefficient for O 2 could be 6½.
C4H10 + 6½O2 → 4CO2 + 5H2O

Now double all the coefficients and the equation is done.
2C4H10 + 13O2 → 8CO2 + 10H2O
PROPORTIONS METHOD:
This method should only be done when the equation is too hard to do by inspection. Always start out doing the
equation by inspection first. If after several minutes of not getting anywhere, use the proportions method.
Example #1: Balance Cu + HNO3 → Cu(NO3)2 + NO + H2O
STEPS:
1) Place variables in front of each substance to represent the coefficients.
a Cu + b HNO3 → c Cu(NO3)2 + d NO + e H2O
2) Since the Law of Conservation of Matter tells us that the same number of atoms of each element must be
found on both sides of the equation, write an equation, using the coefficient variables and subscripts, for
each element so that the number of reactant atoms are equal to the number of product atoms.
Cu
H
N
O
a=c
b = 2e
b = 2c + d
3b = 6c + d + e
3) Take the longest equation, and by substitution get the equation to be a proportion containing only two
different variables. This is the step that will differ from one person to another simply because of the
choice of variables you use—as long as you can get two variables by using correct algebra it will work.
Since b = 2e,
Then: 2e = 2c + d
2e − 2c = d
Then:
3(2e) = 6c + (2e − 2c) + e
6e = 4c + 3e
3e = 4c
e 4

c 3
4) From this proportion, e = 4 and c = 3. Now you can find the other variables using the equations you found
for each element.
a=c
→
a=3
b = 2e
→
b = 2(4) = 8
d = 2e − 2c
→
d = 2(4) − 2(3) = 2
5) Now place these values in the chemical equation for the coefficient variables and you are done.
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
Example #2: Balance HIO3 + FeI2 + HCl → FeCl3 + ICl + H2O
a HIO3 + b FeI2 + c HCl → d FeCl3 + e ICl + f H2O
H
I
O
Fe
Cl
a + c = 2f
a + 2b = e
3a = f
b=d
c = 3d + e
a + c = 2f
a + 2b = e
→
→
a + c = 2(3a)
a + 2d = e
c = 3d + e
→
5a = 3d + (a + 2d)
5a = 5d + a
4a = 5d
a 5

d 4
b=d
c = 5a
→
→
c = 5(5)
→
e = a + 2d
f = 3a
→
→
e = 5 + 2(4)
f = 3(5)
→
→
→
c = 5a
a=5
b=4
c = 25
d=4
e = 13
f = 15
5HIO3 + 4FeI2 + 25HCl → 4FeCl3 + 13ICl + 15H2O