So, how does the balancer work (and how can you do the same thing in an exam)?

The following method follows the way (at least in principle) that the JavaScript programme solves equations:

Take the equation:

Cu + HNO_{3} -> Cu(NO_{3})_{2} + NO + H_{2}O.

First put unknown coefficients on all formulae EXCEPT the first, leave this at 1.

Cu + a HNO_{3} -> b Cu(NO_{3})_{2} + c NO + d H_{2}O.

Then compute the individual atom relationships such that

reactant atoms = product atoms:

Cu | + a HNO_{3} |
--> b
Cu(NO_{3})_{2} |
+ c NO | + d H_{2}O | |

Cu atoms: | 1 | = b | |||

H atoms | a | = | 2 d | ||

N atoms | a | = 2 b | c | ||

O atoms | 3 a | = 6 b | c | d |

So 4 equations can be written in the 4 unknowns:

I:1 = b

II: a = 2d

III: a = 2b + c

IV: 3a = 6b + c + d

Now all that is needed is to solve for the variables a, b, c, and d.

Since b = 1 (I) and a = 2d (III),

substitute for b in III to give (V) a = 2 + c

and for b and d in (IV) to give (VI) 3a = 6 + c + a/2 (or 5a/2 = 6 + c)

From V, substitute c = a - 2 into VI:

5a/2 = 6 + a - 2 (or 3a/2 = 4)

From which a = 8/3.

Now, working backwards: c = a - 2 = 8/3 - 2 = 2/3

and finally: d = a/2 = 4/3.

So the 5 coefficients (in order) are:

1, 8/3, 1, 2/3 and 4/3.

To make these whole numbers, multiply throughout by three to get:

3, 8, 3, 2, and 4.

The balanced equation is therefore:

3 Cu + 8 HNO_{3} -> 3 Cu(NO_{3})_{2} + 2 NO + 4 H_{2}O.

So now you know!

page upkeep: Dave Woodcock.

I would appreciate comments and additions by e mail.