Answer
See solution
Work Step by Step
The following vectors list the numbers of atoms of
potassium (K), manganese (Mn), oxygen (O), sulfur (S), and hydrogen (H):
2
\[
x_{1} \cdot\left[\begin{array}{l}
1 \\
1 \\
4 \\
0 \\
0
\end{array}\right]+x_{2} \cdot\left[\begin{array}{l}
0 \\
1 \\
4 \\
1 \\
0
\end{array}\right]+x_{3} \cdot\left[\begin{array}{l}
0 \\
0 \\
1 \\
0 \\
2
\end{array}\right]=x_{4} \cdot\left[\begin{array}{l}
0 \\
1 \\
2 \\
0 \\
0
\end{array}\right] x_{5} \cdot\left[\begin{array}{l}
0 \\
0 \\
4 \\
1 \\
0
\end{array}\right]+x_{6} \cdot\left[\begin{array}{l}
0 \\
0 \\
4 \\
1 \\
2
\end{array}\right]
\]
The coefficients in the equation
\[
x_{1} \cdot K M n O_{4}+x_{2} \cdot M n S O_{4}+x_{3} \cdot H_{2} O \rightarrow x_{4} \cdot M n O_{2}+x_{5} \cdot k_{2} S O_{4}+x_{6} \cdot H_{2} S O_{4}
\]
Satisfy left side equation
3
\[
\left[\begin{array}{ccccccc}
1 & 0 & 0 & 0 & -2 & 0 & 0 \\
1 & 1 & 0 & -1 & 0 & 0 & 0 \\
4 & 4 & 1 & -2 & -4 & -4 & 0 \\
0 & 1 & 0 & 0 & -1 & -1 & 0 \\
0 & 0 & 2 & 0 & 0 & -2 & 0
\end{array}\right]
\]
Move the right terms to the left side and reduce in the augmented matrix form
4
\[
\left[\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 & 0 & -1.5 & 0 \\
0 & 0 & 1 & 0 & 0 & -1.0 & 0 \\
0 & 0 & 0 & 1 & 0 & -2.5 & 0 \\
0 & 0 & 0 & 0 & & -0.5 & 0
\end{array}\right]
\]
After some row operations
5
The general solution is
\[
\begin{array}{c}
x_{1}=x_{6} \\
x_{2}=(1.5) x_{6} \\
x_{3}=x_{6} \\
x_{4}=(2.5) x-6 \\
x_{5}=.5 x_{6}
\end{array}
\]
and $x_{6}$ is free. Take $x_{6}=2 .$ Then $x_{1}=x_{3}=2,$ and $x_{2}=3, x_{4}=$
$5,$ and $x_{5}=1 .$ The balanced equation is
\[
2 . K M n O_{4}+3 . M n S O_{4}+2 . H_{2} O \rightarrow 5 . M n O_{2}+k_{2} S O_{4}+2 . H_{2} S O_{4}
\]
