Answer
B2S3+6H2O=2H3BO3+3H2S
Work Step by Step
\[
\begin{array}{l}
B_{2} S_{3}=\left[\begin{array}{l}
2 \\
3 \\
0 \\
0
\end{array}\right] ; H_{2} O=\left[\begin{array}{l}
0 \\
0 \\
2 \\
1
\end{array}\right] ; H_{3} B O_{3}=\left[\begin{array}{l}
1 \\
0 \\
3 \\
3
\end{array}\right] ; H_{2} S= \\
{\left[\begin{array}{l}
0 \\
1 \\
2 \\
0
\end{array}\right]}
\end{array}
\]
A systematic method for balancing the chemical equation is to set up a vector equation that describe the number of atoms of each type present in a reaction. since given equation involves 4 types of atoms. Construct a vector equation for each reactant and product.
\[
x_{1}\left[\begin{array}{l}
2 \\
3 \\
0 \\
0
\end{array}\right]+x_{2}\left[\begin{array}{l}
0 \\
0 \\
2 \\
1
\end{array}\right]=x_{3}\left[\begin{array}{l}
1 \\
0 \\
3 \\
3
\end{array}\right]+x_{4}\left[\begin{array}{l}
0 \\
1 \\
2 \\
0
\end{array}\right]
\]
To balance given equation, the coefficient must achieve
this.
\[
x_{1}\left[\begin{array}{l}
2 \\
3 \\
0 \\
0
\end{array}\right]+x_{2}\left[\begin{array}{l}
0 \\
0 \\
2 \\
1
\end{array}\right]+x_{3}\left[\begin{array}{c}
-1 \\
0 \\
-3 \\
-3
\end{array}\right]+x_{4}\left[\begin{array}{c}
0 \\
-1 \\
-2 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0
\end{array}\right]
\]
\[
\left[\begin{array}{cccc}
2 & 0 & -1 & 0 \\
3 & 0 & 0 & -1 \\
0 & 2 & -3 & -2 \\
0 & 1 & -3 & 0
\end{array}\right] \sim\left[\begin{array}{cccc}
0 & 0 & -3 & 2 \\
1 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 \\
0 & 1 & -3 & 0
\end{array}\right]
\]
Row reduce the augmented matrix.
\[
\begin{array}{l}
x_{1}=\frac{1}{3} x_{4} \\
x_{2}=2 x_{4} \\
x_{3}=\frac{2}{3} x_{4} \text { with } x_{4} \text { free }
\end{array}
\]
Row reduction of augmented matrix leads to these
general solution.
\[
x_{1}=1, x_{2}=6, x_{3}=2, x_{4}=3
\]
To avoid fraction take $x_{4}=3$ and find $x_{1}, x_{2}, x_{3}$
\[
B_{2} S_{3}+6 H_{2} O=2 H_{3} B O_{3}+3 H_{2} S
\]
