Talk:Chemical equation

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But I ask someone to add some better method for equation balancing since what is there although is kinda practical reasoning, it's not so suitable for complex cases and also takes a lot of text to explain(someone should clean that up too).

But what I really want is to suggest a description of linear systems balancing, which I find much more effective and elegant. I will give an example:

K₄Fe(CN)₆ + H₂SO₄ + H₂O → K₂SO₄ + FeSO₄ + (NH₄)₂SO₄ + CO

The method consists in the following:

1. Assign variables to each coefficient:

a K₄Fe(CN)₆ + b H₂SO₄ + c H₂O → d K₂SO₄ + e FeSO₄ + f (NH₄)₂SO₄ + g CO

2. We must have the same quantities of each atom in each side of the equation. So, for each element, count its atoms and equal both sides:

K: 4a = 2d

Fe: 1a = 1e

C: 6a = g

N: 6a = 2f

H: 2b+2c = 8f

O: 4b+c = 4d+4e+4f+g

3.

d=2a

e=a

g=6a

f=3a

b=6a

c=6a

which means that we have all coefficients depending on a parameter a, just choose a=1(a number that will make all of them small whole numbers) and you'll have:

a=1 b=6 c=6 d=2 e=1 f=3 g=6

4. And the balanced equation at last:

K₄Fe(CN)₆ + 6 H₂SO₄ + 6 H₂O → 2 K₂SO₄ + FeSO₄ + 3 (NH₄)₂SO₄ + 6 CO

One could argue that this method requires much more work than the other, for that reason I will show that combining both can lead to a very efficient and quick way of balancing(although for very complex cases I would keep this one in the pocket):

1. Identify elements which occur in one compound in each member(this is very usual)

2. Start with the one among those which has a big index(this will help to keep working with integers), and assign a variable, let's say a.

a K₄Fe(CN)₆ + H₂SO₄ + H₂O → K₂SO₄ + FeSO₄ + (NH₄)₂SO₄ + CO

3. Well, K₂SO₄ has to be 2a(because of K), and also, FeSO₄ has to be 1a(because of Fe), CO has to be 6a(because of C) and (NH₄)₂SO₄ has to be 3a(because of N). Well, this takes out the first four equations of the system! We already now that, whatever the coefficients are, those proportions must hold:

a K₄Fe(CN)₆ + H₂SO₄ + H₂O → 2a K₂SO₄ + a FeSO₄ + 3a (NH₄)₂SO₄ + 6a CO

4. We can continue by writing the equations now(and having simpler problem to solve) or, in this particular case(although not so particular) we could continue by noticing that adding the Sulfurs we get 6a for H₂SO₄ and finally by adding the hydrogens(or the oxygens) we get the lasting 6a for H₂SO₄.

5. Again, having a convienient value for a(in this case 1 will do, but if a gets fractionary values in the other coefficients you will like to cancel the denominators) we get the result:

K₄Fe(CN)₆ + 6 H₂SO₄ + 6 H₂O → 2 K₂SO₄ + FeSO₄ + 3 (NH₄)₂SO₄ + 6 CO

Rend 03:06, 10 August 2005 (UTC)

I added it as I described it here, I think it need to be better written although. Rend 04:21, 19 December 2005 (UTC)

I don't know if this helps anyone but I thought I would mention that the reverse arrows display as boxes in MSN Explorer 9.2 64.114.88.154 23:14, 30 March 2006 (UTC)

i hate u sharna !!!