So, how does the balancer work (and how can you do the same thing in an exam)?
Take the equation:
Cu + HNO3 -> Cu(NO3)2 + NO + H2O.
First put unknown coefficients on all formulae EXCEPT the first, leave this at 1.
Cu + a HNO3 -> b Cu(NO3)2 + c NO + d H2O.
Then compute the individual atom relationships such that
reactant atoms = product atoms:
|Cu||+ a HNO3||--> b Cu(NO3)2||+ c NO||+ d H2O|
|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 HNO3 -> 3 Cu(NO3)2 + 2 NO + 4 H2O.
So now you know!