In my high school chem class, we were taught the "Oxidation State" method. Going with quad's example ... KMnO4 + HCl ---> KCl + MnCl2 + H2O + Cl2 We use our chemistry knowledge to figure out that only Mn and Cl are changing oxidation state here. (Oxidation state = signed valence, from a math viewpoint. Usually, H = K = +1, O = -2, Cl = -1. Stable compounds have a total of 0, such as HCl, KCl, and H2O.) What's happening in this reaction is that Mn has an oxidation state of +7 in KMnO4 (potassium permanganate, an oxidizing agent), and it drops to +2 in MnCl2. The Mn is reduced, and some of the chloride (state -1) is oxidized to 0 (plain Chlorine, as Cl2). In theory, these represent electrons being shuffled around between atoms, filling completed shells. Fie. The deltas are Mn: +7 -> +2 and Cl: -1 -> 0. So we expect 1 "molecule" of KMnO4 to balance 5 atoms of Cl (as 2.5 * Cl2). Our first cut has 1 KMnO4 + ? HCl ---> ? KCl + ? MnCl2 + ? H2O + 2.5 Cl2 We can double this to clear the fraction; this is a matter of taste. 2 KMnO4 + ? HCl ---> ? KCl + ? MnCl2 + ? H2O + 5 Cl2 Now we fill in the obvious other coefficients, balancing K, Mn, and O from LHS to RHS: 2 KMnO4 + ? HCl ---> 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2 The HCl coefficient is the only thing left, and it must be 16 to make the H from H2O work. Then we check the Cl balance, and we're done. I've admired this method for its "power": it lets you do stuff in your head that would otherwise be hard. It's subject to criticism on both chemical and mathematical grounds: Reactions don't go to completion, how do you know what products to expect, why doesn't the Cl2 react with the water, ... . For mathematicians, a lot of reactions don't have a unique solution, and how do we know to look at Mn and Cl as the change agents? Why doesn't KMnO4 in water make peroxide (H2O2) or oxygen (O2)? There's so much in chemistry that seems vague & ambiguous. Rich ---------- Quoting quad <quadricode@gmail.com>:
On Sun, Apr 24, 2011 at 2:22 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Marc:
It's been a while (nearly 50 years) since I've done this; could you give an example of what you mean?
He means, given a chemical "equation", determine the "coefficients" of each molecule.
Example: Given
KMnO4 + HCl ---> KCl + MnCl2 + H2O + Cl2
determine a, b, ..., f so that
a*KMnO4 + b*HCl ---> c*KCl + d*MnCl2 + e*H2O + f*Cl2
"balances", i.e., there are equal numbers of each type of atom on each side. The balanced equation here is
2 KMnO4 + 16 HCl ---> 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2
because we have equal numbers of each atom. Look at O for example. On the left, we have 2*4 = 8 O, on the right, we have 8*1 O.
Anyway, I think you get it. :)
- Robert
At 01:13 PM 4/24/2011, Marc LeBrun wrote:
Can anyone come up with a nice way to balance chemical equations manually?
All the web seems to offer is either vague "fiddle around until it works" or the nuclear option "translate into a simultaneous linear system and solve".
Is there anything in between? It need not be theoretically optimal, just easy to apply by hand to small solvable cases.
I'm imagining a well-defined procedure repeatedly "adjusting" coefficients until "done", then dividing out their common factor, perhaps akin to an n-D raster line drawing algorithm that somehow manages to hill climb onto a scaled solution.
It should be more clever than, say, mindlessly trying all the possible cases in some fixed order, yet stay grounded in the problem domain.
It might even be "morally equivalent" to Gaussian elimination but performed directly on the chemical equations. Longhand division is kind of like this. There's a little eyeballing and maybe even some backtracking estimating the digits, but it's a reasonably effective way to arrive at the answer by hand. Crunching determinants for simple chemistry is analogous to using Newton's method on everyday division problems.
Any ideas?
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