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This question has been intriguing me since 10th or 11th grade. The teacher just told us about it but didn't get into the details of the why. Recently I asked a biochemist but couldn't get an answer. So I bring it here.

$$\ce{CH3OH + O2 -> CO2 + H2O}$$

is the complete combustion equation of methanol. But the problem is it can't be balanced. What is the practical implication, and significance of this fact?

Side question:

What is the mathematical reasoning behind the inability to balance the equation?

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    $\begingroup$ Wait, what do you mean it cannot be balanced? $\endgroup$ – LordStryker Jan 29 '14 at 21:58
  • $\begingroup$ Maybe it was some other equation. Just can't recall it, I'll update the question. $\endgroup$ – Bleeding Fingers Jan 29 '14 at 22:02
  • $\begingroup$ Please do. I'm very curious now. $\endgroup$ – LordStryker Jan 29 '14 at 22:03
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    $\begingroup$ I'm pretty sure every single valid chemical equation balances, as long of course as you know all the products. You could conceivably use a reaction as an analog computer (a very cumbersome one!) to solve a linear system of equations. If nature does it, then math must do it. $\endgroup$ – Nicolau Saker Neto Jan 29 '14 at 22:41
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    $\begingroup$ This question has awoken the fire of linear algebra in me! I believe it is possible to prove that any system of equations resulting from a valid reaction has $n$ variables and $n-1$ equations, i.e. the system is always underdetermined by one equation and therefore has infinite solutions, all of which lie in a line in $\mathbb{R}^n$ sand are multiples of eachother by some arbitrary real number. Let me try formalizing the argument. Unfortunately I'm kind of tired, so I can't guarantee I'll be able to wade through the notation properly. $\endgroup$ – Nicolau Saker Neto Jan 29 '14 at 23:09
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Here is the balanced eqn...

$$\ce{2 CH3OH + 3 O2 → 2 CO2 + 4 H2O}$$

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    $\begingroup$ While i was typing the the mathjaxed equation, you were faster submitting. So I beautify yours :-D $\endgroup$ – Klaus-Dieter Warzecha Jan 29 '14 at 22:02
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    $\begingroup$ @KlausWarzecha Can you beautify me next?? $\endgroup$ – LordStryker Jan 29 '14 at 22:06
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    $\begingroup$ @LordStryker I might be good, but not THAT good :-P $\endgroup$ – Klaus-Dieter Warzecha Jan 29 '14 at 22:10
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    $\begingroup$ beatify or beautify? $\endgroup$ – ron Jan 29 '14 at 22:23
  • $\begingroup$ @ron Blessed are the cheesemakers. $\endgroup$ – Klaus-Dieter Warzecha Jan 29 '14 at 22:53
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Mass is conserved, charge is conserved, ignore spectator ions until the end. If it is a redox reaction, begin by listing each active atom's reduction or oxidation, e.g.,

Mn(+7) + 5e to Mn(+2)

Cr(+3) - 3e to Cr(+6)

and find the least common denominator (here, 15). That gives you the primary stoichiometry, here 5Cr(+3) (going to chromate) with 3Mn(+7) (coming from permanganate). Add the other atoms, slop in H2O, OH-, or H+, add back the spectator ions. Confirm the same number of each kind of atom on both sides. Try it out,

CuSCN + KIO3 + HCl to CuSO4 + KCl + HCN + ICl + H2O

[Cr(N2H4CO)6]4[Cr(CN)6]3 + KMnO4 + H2SO4 to K2Cr2O7 + MnSO4 +CO2 + KNO3 + K2SO4+ H2O

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