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Question
Balance the following chemical equation (Find a, b, c, d, e and f)

a(Cr_2 O_7)^(2-) + b(H)^(+) + c(H_2 SO_3) -> d(Cr)^(3+) + e(H_2 O) + f(HSO_4)^(-)

I'm confused... My solutions don't make sense, I am getting negative numbers..?

I'm not sure about the the number after the ^ mean.
With the minus sign and plus sign after the number, that may be the charge,
but I'm not sure.  Perhaps that would add the equation -2a + b = 3d - f.
See bottom for the two cases I found that work for this.

On the left, we have 2a Cr, b+2c H, 7a+3c O, c S.
On the right, we have d Cr, 2e+f H, e+4f O, and f S.

That gives us the equations [1] 2a = d, [2] b+2c = 2e+f, [3] 7a+3c = e+4f, and [4] c=f.

Looking at [4], we can see that gives c can be substituted for f.
Putting that in the 1st three equations gives [1] 2a = d, [2] b+2c = 2e+c => b+c = 2e,
and [3] 7a+3c = e+4c => 7a = e+c.

That is, [1] 2a = d, [2] b+c = 2e, and [3] 7a = e+c.

Whatever a is chosen, it must be even, for 2a = d, and d only occurs in [1].
Solving [2] and [3] for c gives [2] c = 2e-b and [3] c = 7a-e.

Equating these two gives 2e-b = 7a-e.  Putting e on the left and a and b on the right,
we have 3e = 7a + b.

Since we have integer solutions, lets try them out.
If e=1, that makes a=0 and b=3, but they need to be positive integers.
If e=2, there is no way this could occur.

If e=3, we could have a=1 and b=2 since 7 + 2(1) = 7+2 = 9.
Since a=1, d=2a says d=2.
For equation [2] c = 2e-b. this says c = 2*3 - 2 = 4.
We already know that f=c, so f is 4 as well.

This means we have a,b,c,d,e,f as 1,2,4,2,3,4.

Checking the equation, for [1] we get  2(1) = 2, and that is true.
For [2] we get 2+2(4) = 2(3)+4 => 2+8 = 6+4 => 10=10, and that is true
For [3], we get 7+3(4) = 3+4(4) => 7+12 = 3+16 => 19 = 19, ane that is true.
For [4], we get 4 = 4, and that is true.

Since all equations are satisfied, this works.

Note that since there were only 4 equations and 6 variables, there are multiple solutions, but in chemistry, the simplest one is used.

We could also have a,b,c,d,e,f as 1,5,3,2,4,3.  On the math side, that reminds me of 153 and 243.  153 is 1^3 + 5^3 + 3^3 since that is 1 + 125 + 27 = 153 and 243 = 2^5, since that is
3*3*3*3*3.

Back to the problem, I found 48 solutions that worked with a being between 1 and 5.
They are
a   b   c   d   e   f
1   11   1   2   6   1
1   8   2   2   5   2
1   5   3   2   4   3
1   2   4   2   3   4
2   25   1   4   13   1
2   19   3   4   11   3
2   13   5   4   9   5
2   7   7   4   7   7
2   1   9   4   5   9
3   39   1   6   20   1
3   36   2   6   19   2
3   30   4   6   17   4
3   27   5   6   16   5
3   21   7   6   14   7
3   18   8   6   13   8
3   12   10   6   11   10
3   9   11   6   10   11
3   3   13   6   8   13
4   53   1   8   27   1
4   47   3   8   25   3
4   41   5   8   23   5
4   35   7   8   21   7
4   29   9   8   19   9
4   23   11   8   17   11
4   17   13   8   15   13
4   11   15   8   13   15
4   5   17   8   11   17
5   67   1   10   34   1
5   64   2   10   33   2
5   61   3   10   32   3
5   58   4   10   31   4
5   52   6   10   29   6
5   49   7   10   28   7
5   46   8   10   27   8
5   43   9   10   26   9
5   37   11   10   24   11
5   34   12   10   23   12
5   31   13   10   22   13
5   28   14   10   21   14
5   22   16   10   19   16
5   19   17   10   18   17
5   16   18   10   17   18
5   13   19   10   16   19
5   7   21   10   14   21
5   4   22   10   13   22
5   1   23   10   12   23

The one case is a,b,c,d,e,f as 1, 5, 3, 2, 4, 3.

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#### Scott A Wilson

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