Uploaded by Peter Goodwill


Let's try an algebraic method for
H3BO3 → H4B6O11 + H2O
It can be rather easily balanced by inspection, but let's try a more systematic approach.
What does 'balanced' mean? It means that for every element, there is the same number of atoms on both
sides of the reaction equation. Our reaction has three coefficients a, b and c:
aH3BO3 → bH4B6O11 + cH2O
'Balanced' means that there is exactly the same number of atoms of each element on both sides of the
equation. Using coefficients a, b and c we can tell that we have 1×a atoms of boron on the left (one atom
per each H3BO3 molecule), and 6×b + 0×c on the right (6 atoms of boron per each H4B6O11 molecule and
no boron in water). This gives us the following equation:
1×a = 6×b + 0×c
We can write similar equations for all elements - hydrogen:
3×a = 4×b + 2×c
and oxygen:
3×a = 11×b + c
As there are no free terms in this set of equations, it has a trivial solution (a = b = c = 0) which we are
not interested in. We have three equations, and three unknowns - nothing particularly difficult to solve.
Quite often you will end with many more equations and many more unknowns. Such equation sets is not
a thing that you may want to solve manually, although when balancing chemical equations in most cases
it can be done relatively easy, as most equations don't contain all unknowns. In this case we have a very
simple equation a = 6×b that we can use to substitute 6×b for a in the second and third equation to get:
18×b = 4×b + 2×c
18×b = 11×b + c
After some rearranging:
7×b = c
7×b = c
Both equations are identical. In algebra it usually means that the set of equations doesn't have a unique
solution, but in the case of chemical equations we have one additional information - all coefficients must
be an integer and they must be the smallest ones. To find them we can assume one of the coefficients to
be 1:
If so
and indeed
6H3BO3 → H4B6O11 + 7H2O
is the balanced reaction equation.
This first example doesn't look convincing - why do we have to solve a set of equations when the reaction
equation can be easily balanced by other means? Good point - but what if the reaction can be not easily
P2I4 + P4 + H2O → PH4I + H3PO4
Try for a moment. Looks easy but soon gets surprisingly hard and the coefficients become pretty high,
which makes you wonder if you have not made some mistake (*see some balancing hints at the bottom of
the page). What about the general, algebraic method?
We need five coefficients, and there are only four equations (one for each element present) - but it
shouldn't bother us, as we know that we have additional information that works as an additional equation.
aP2I4 + bP4 + cH2O → dPH4I + eH3PO4
Setting up equations:
P: 2×a + 4×b = d + e
I: 4×a = d
H: 2×c = 4×d + 3×e
O: c = 4×e
Balances for iodine and oxygen make this set look much easier than expected. c = 4×e and d = 4×a are
substitutions that we are about to use to reduce number of unknowns:
1st equation: 2×a + 4×b = 4×a + e
3rd equation: 8×e = 4×d + 3×e
So we have now after some cancelling:
4×b = 2×a + e
4×a = d
5×e = 4×d
c = 4×e
For someone fluent in algebra it is obvious that we have already finished - it is now enough to assume
that one of the variables equals 1 to calculate values of all others. Assuming a = 1 and simply
substituting calculated values we have:
b = 13/10
c = 64/5
e = 16/5
These are hardly integer, but all we have to do is to find the smallest common denominator to have a list
of integer coefficients in numerators. In this case the smallest common denominator is 10, so if we
multiply all numbers by 10 we get:
a = 10
b = 13
c = 128
d = 40
e = 32
And you may check that these are the correct coefficients. Imagine finding them by the inspection
It's also easy to use the algebraic method to balance redox reactions with charged species. Let's try it for
aCr2O72- + bH+ + cFe2+ → dCr3+ + eH2O + fFe3+
What equations do we have? Four balances of atoms:
Cr: 2a = d
O: 7a = e
H: b = 2e
Fe: c = f
But that's not enough to balance the equation - we have six coefficients and four equations. For the
algebraic method we cannot have one equation less than variables - so we are still one equation short.
However, charge has to be balanced as well, and that will give us the last equation needed to balance the
-2a + b + 2c = 3d + 3f
(Note that the sign of the coefficients in the last equation depends on the charge sign). We are ready to
solve. First of all, we know that
e = 7a
b = 14a
Let's get rid of b and d in the charge balance equation:
-2a + 14a + 2c = 6a + 3f
Sorting, and using c = f:
c = 6a
That's almost ready. Let's put a = 1 and calculate all other coefficients simply using already known
b = 14
and the balanced equation takes form:
Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
That's all. Not that hard.
Probably the most important characteristics of the algebraic method is that - contrary to the inspection
method - is guaranteed to give you an answer. If the reaction can be balanced, you will find coefficients.
If the reaction can't be balanced - you will find it out seeing that there are more unknowns than
independent equations (remember - one more is not a problem), or that equations are contradictory.
With the inspection method you will never have a proof that the equation can't be balanced.
There is an additional reason for this method to be important. Computers are very effective in solving
sets of simultaneous equations, for example using a method called Gauss elimination. What is difficult for
humans is a perfect task for the number cruncher built in every processor. That's why EBAS is able to
balance the so-called Blakley equation (20 unknowns in 19 equations) in a blink of an eye on a 10 year
old PC.
Note that there are - although rare - cases, when a reaction occurring in reality cannot be balanced with
the algebraic approach.
Balancing hints for P2I4 + P4 + H2O → PH4I + H3PO4 reaction:
Contrary to what balancing by inspection rules say, start with balancing oxygen from phosphoric acid,
then balance hydrogen, iodine, and finally phosphorus. Leaving oxygen and hydrogen to the end is asking
for trouble.
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