Strikes me, MrT, that you have to be wrong because everything is multiplied by two, giving no overall change at all. In particular, there's 6O on the left and 4O on the right.
Now, I can't really help you in a useful chemical way because I'm not a chemistry student. I could teach you the mathsy way of doing it but I don't know if that really helps.
Anyway, here goes...
It should be:
3Co + 2HNO3 + 6HCl ----> 3CoCl2 + 2NO + 4H2O
I've checked this and it works.
Now, how?
Mathematically, this is expressing a set of 5 simultaneous equations (for each element) with 6 unknowns (how many of each compound) - though you don't need to know this. So what we do is go through this chemical equation element by element, counting how many times each element appears on each side, and write the numbers we are looking for (how many of each compound) as letters: a, b, c, d, e, f I chose.
This got me to the following:
- aCo = dCo (by comparing the first compound on each side)
- (b+c)H = 2fH (where H2 has become "count H twice" - and I got this one wrong first time I tried)
- bN = eN (comparing the second compound on each side)
- 3bO = (e+f)O (O3 is read as 3 lots of O )
- cCl = 2dCl
What does this tell us? Now I jsut removed all the elements!
a = d
b = e
c = 2d
3b = e + f
b + c = 2f
And now it's just an exercise in algebra. After some fiddling I ended up with:
2c = 4d = 4a = 6b = 6e = 3f
So at least all the numbers we are looking for are linked. Now, it was a question of trial and error. The final condition on a, b, c, d, e and f is that they're all whole numbers, and as small as possible, so I just tried b = e = 1 and this failed, so I then tried b = e = 2. This worked, and is hence the answer.
