# Balancing Chemical Equations with Linear Algebra

1. Feb 1, 2010

### Squall

1. The problem statement, all variables and given/known data
Balance the following Chemical Equation
(x1)KMnO4+(x2)MnSO4+(x3)H2O->(x4)MnO2+(x5)K2SO4+(x6)H2SO4

2. Relevant equations
Here is the matrix I came up with but it seems to be flawed but I am unable to see what i am doing wrong

1 0 0 0 -2 0 (K)
0 1 0 -1 0 0 (Mn)
4 4 1 -2 -4 -4 (O)
0 1 0 0 -1 -1 (S)
0 0 2 0 0 -2 (H)
Each column corresponds to the amount of each element in each compound.

This is the reduced row echelon form of the matrix
1 0 0 0 0 -1/3
0 1 0 0 0-7/6
0 0 1 0 0 -1
0 0 0 1 0-7/6
0 0 0 0 1 -1/6

I seem to be missing the free variable I need to solve this problem

3. The attempt at a solution

The answer i come up with is
X1=(1/3)
X2=(7/6)
X3=(1)
X4=(7/6)
X5=(1)

and here is the problem for I only get five answers and not the six that i need to balance the equation. Can anyone point me in a direction or say what I am doing wrong any help will be appreciated thank you.

2. May 18, 2011

### yazooj

you need to add a column of zeros on the right of your matrix

3. Feb 8, 2012

### muntazimabbas

Balance is made by taking x6 as a free variable. This variable is free because in the Matrix, x6 is the position where no pivot element is present, i.e. column of x6 is not a pivot column. So compute the remaining variables in the form of x6 and then multiply all the values of x1 ... x5 by the LCM of the denominators of the fractions.

In my view the answer will look like (for example if I suppose the obtained Echelon form is correct, i.e. This is the reduced row echelon form of the matrix
1 0 0 0 0 -1/3
0 1 0 0 0-7/6
0 0 1 0 0 -1
0 0 0 1 0-7/6
0 0 0 0 1 -1/6)

X1=(1/3)t
X2=(7/6)t
X3=(1)t
X4=(7/6)t
X5=(1/6) t
x6=t, where t is in Set of Real Numbers

OR

X1=(2)t
X2=(7)t
X3=(6)t
X4=(7)t
X5=(1) t
x6=(6)t, where t is in Set of Real Numbers

4. Feb 8, 2012