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I tried balancing this chemical equation with a system of equations, but the answer doesn't seem to map well. I get a negative coefficient which is prohibitive in this equations. How do I interpret the answer?

$Al(OH)_3 + H_2SO_4 \rightarrow Al_2(SO_4)_3 + H_2O ${{1,0,2,0},{3,4,12,1},{3,2,0,2},{0,1,3,0}}


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up vote 4 down vote accepted

Let $x$ be the number of $Al(OH)_3$; $y$ the number of $H_2SO_4$; $z$ the number of $AL_2(SO_4)_3$, and $w$ the number of $H_2O$. Looking at the number of $Al$, you get $x = 2z$. Looking at $O$, you get $3x + 4y = 12z + w$. Looking at $H$ you get $3x + 2y = 2w$; and looking at $S$ you get $y = 3z$. That looks like what you are getting from Wolfram, except you have the wrong signs for $z$ and $w$; unless you are interpreting the first two entries to represent the "unknowns", and the last two to represent the "solutions". I would translate into equations the usual way.

What you have is the following system of linear equations: $$\begin{array}{rcrcrcrcl} x & & & & -2z & & & = & 0\\ 3x & + & 4y & - & 12z & - & w & = & 0\\ 3x & + & 2y & & & - & 2w & = & 0\\ & & y & - & 3z & & & = & 0 \end{array}$$ This leads (after either some back-substitution from the first and last equations into the second and third, or some easy row reduction) to $x=2z$, $y=3z$, and $6z=w$. Since you only want positive integer solutions, setting $z=1$ gives $x=2$, $y=3$, and $w=6$, yielding the smallest solution: $$2 Al(OH)_3 + 3H_2SO_4 \longrightarrow Al_2(SO_4)_3 + 6H_2O.$$

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Ah thanks that's very helpful :) – Ted Flethuseo Jan 21 '11 at 16:53

Why don't you just use the actual chemical equation as input?

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Ah very useful, thanks. I still however want to do it through a system of equations, I am only allowed to use a calculator on the test. – Ted Flethuseo Jan 21 '11 at 16:54

A chemical balance will always admit the solution with all zero coefficients: if you put nothing in, you get nothing out. Or, if you double the amount you put in, you will get double the output.

In linear algebra language, this means that your solution cannot be unique. The set of admissible solutions must form a linear subspace.

The set of admissible solutions will be the kernel of the matrix you wrote down corresponding to the chemical balance (incidentally, I would put negative signs in front of the first two columns since those represent things on the left hand side of the chemical equation and are things that are consumed), and you are looking for a vector in this kernel with all positive coefficients. Therefore it is in fact necessary for the row-reduced matrix to have negative entries: if the row-reduced matrix has only positive value entries, then it is impossible to have a vector with positive entries in its kernel (since a sum of positive numbers must be positive, and not zero).

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Unless I'm misunderstanding, the right side dictates that the ratio of Al to S should be $2:3$, so the coefficients on the left side must be in the ratio $2:3$. Suppose they are 2 and 3. There are now $2\cdot3+3\cdot4=18$ O and $2\cdot3+3\cdot2=12$ H on the left. On the right, the coefficient of the first term must be 1 (since we used 2 and 3 on the left, there is enough Al and S for only 1), which uses 12 O, leaving 6 O and 12 H. Setting the coefficient of the last term to 6 matches up with 6 O and 12 H.

Looking at the results from your row-reduction on WolframAlpha, I'm thinking that the $-1$ is because the matrix form is working as if that term were on the same side as the 2 and 3, but on the opposite side, it would be a positive $1$.

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