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Balance chemical equations without trial and error?

Thanks Ben, the other answer was helpful... but is it possible to do it when, say, five variables are involved?

E.g.: $\textrm{Na}_2\textrm{CO}_3 + \textrm{HCl} \to \textrm{NaCl} + \textrm{H}_2\textrm{O} + \textrm{CO}_2$

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marked as duplicate by ShreevatsaR, joriki, Ilmari Karonen, J. M., Asaf Karagila Nov 16 '11 at 8:14

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Please make the question self-contained. In this case, add a link to the "other answer" that was helpful... –  Srivatsan Nov 16 '11 at 4:19
    
If you recognize $\textrm{H}_2\textrm{O} + \textrm{CO}_2$ as $\textrm{H}_2\textrm{CO}_3$ you can see you need two $\textrm{HCl}$ to balance the hydrogen and you will get there. But Michael Lugo's approach will work all the time. –  Ross Millikan Nov 16 '11 at 4:39
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1 Answer 1

Write coefficients to be solved for in front of each of the chemical species which occur here:

$$ a \textrm{Na}_2\textrm{CO}_3 + b \textrm{HCl} \to c \textrm{NaCl} + d \textrm{H}_2\textrm{O} + e \textrm{CO}_2$$

and then set up one equation for each element involved in the reaction, making sure that the same number of atoms of that element are on each side of the equation. In this case you get $2a = c$ (from Na), $a = e$ (C), $3a = d + 2e$ (O), $b = 2d$ (H), and $b = c$ (Cl).

We can try to write everything in terms of $a$. We get $c = 2a$ and $e = a$ right away. From $b = c$ we get $b = 2a$, and from $b = 2d$ we get $d = a$. Finally we need to check that $3a = d + 2e$; this is true.

So we let $a = 1$ (in general you want to pick $a$ to be the smallest number it can be and still have all the other coefficients work out to be integers). The result is the balanced equation

$$ \textrm{Na}_2\textrm{CO}_3 + 2 \textrm{HCl} \to 2 \textrm{NaCl} + \textrm{H}_2\textrm{O} + \textrm{CO}_2.$$

(It's also possible to write the system of equations here in terms of matrices, but I find it hard to imagine chemical examples where that would really be necessary.)

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