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The number of atoms of the chemical compound participating in the chemical reaction can be represented in the form of vectors.

The following vectors represent the number of atoms of lead (Pb), nitrogen (N), chromium (Cr), manganese (Mn), and oxygen (O) for the chemical compounds participating in the chemical reaction.

\({\rm{Pb}}{{\rm{N}}_6}:\left[ {\begin{array}{*{20}{c}}1\\6\\0\\0\\0\end{array}} \right]\), \({\rm{CrMn}}{{\rm{O}}_2}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\2\\8\end{array}} \right]\), \({\rm{P}}{{\rm{b}}_3}{{\rm{O}}_4}:\left[ {\begin{array}{*{20}{c}}3\\0\\0\\0\\4\end{array}} \right]\), \({\rm{C}}{{\rm{r}}_2}{{\rm{O}}_3}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\0\\2\\0\\3\end{array}} \right]\), \({\rm{Mn}}{{\rm{O}}_2}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\0\\0\\1\\2\end{array}} \right]\),and \({\rm{NO:}}\left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\\1\end{array}} \right]\)

In the chemical equation, using the vectors, the coefficients of the chemical compound can be determined.

Let the chemical reaction be:

\({x_1} \cdot Pb{N_6} + {x_2} \cdot CrM{n_2}{O_8} \to {x_3} \cdot P{b_3}{O_4} + {x_4} \cdot C{r_2}{O_3} + {x_5} \cdot Mn{O_2} + {x_6} \cdot NO\)

The above reaction must satisfy the equation:

\({x_1}\left[ {\begin{array}{*{20}{c}}1\\6\\0\\0\\0\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\2\\8\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}3\\0\\0\\0\\4\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}0\\0\\2\\0\\3\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}0\\0\\0\\1\\2\end{array}} \right] + {x_6}\left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\\1\end{array}} \right]\)

The vector equation for the chemical compound is used to write the augmented matrix.

Theaugmented matrix for the equation \({x_1}\left[ {\begin{array}{*{20}{c}}1\\6\\0\\0\\0\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\2\\8\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}3\\0\\0\\0\\4\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}0\\0\\2\\0\\3\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}0\\0\\0\\1\\2\end{array}} \right] + {x_6}\left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\\1\end{array}} \right]\) is:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\6&0&0&0&0&{ - 1}&0\\0&1&0&{ - 2}&0&0&0\\0&2&0&0&{ - 1}&0&0\\0&8&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0\end{array}} \right]\)

Row operations do not affect a linear system.

At row 2, multiply row 1 with 6 and subtract it from row 2, i.e., \({R_2} \to {R_2} - 6{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\{6 - 6\left( 1 \right)}&{0 - 0}&{0 - 6\left( { - 3} \right)}&{0 - 0}&{0 - 0}&{ - 1 - 0}&{0 - 0}\\0&1&0&{ - 2}&0&0&0\\0&2&0&0&{ - 1}&0&0\\0&8&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&0&{18}&0&0&{ - 1}&0\\0&1&0&{ - 2}&0&0&0\\0&2&0&0&{ - 1}&0&0\\0&8&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0\end{array}} \right]\)

Interchange the second row and third row, i.e., \({R_2} \leftrightarrow {R_3}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&{18}&0&0&{ - 1}&0\\0&2&0&0&{ - 1}&0&0\\0&8&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0\end{array}} \right]\)

At row 4, multiply row 2 with 2 and subtract it from row 4, i.e., \({R_4} \to {R_4} - 2{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&{18}&0&0&{ - 1}&0\\0&{2 - 2\left( 1 \right)}&{0 - 0}&{0 - 2\left( { - 2} \right)}&{ - 1 - 0}&{0 - 0}&{0 - 0}\\0&8&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&{18}&0&0&{ - 1}&0\\0&0&0&4&{ - 1}&0&0\\0&8&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0\end{array}} \right]\)

At row 5, multiply row 2 with 8 and subtract it from row 5, i.e., \({R_5} \to {R_5} - 8{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&{18}&0&0&{ - 1}&0\\0&0&0&4&{ - 1}&0&0\\0&{8 - 8\left( 1 \right)}&{ - 4 - 0}&{ - 3 - 8\left( { - 2} \right)}&{ - 2 - 0}&{ - 1 - 0}&{0 - 0}\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&{18}&0&0&{ - 1}&0\\0&0&0&4&{ - 1}&0&0\\0&0&{ - 4}&{13}&{ - 2}&{ - 1}&0\end{array}} \right]\)

Divide row 3 by 18, i.e., \({R_3} \to \frac{{{R_3}}}{{18}}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&4&{ - 1}&0&0\\0&0&{ - 4}&{13}&{ - 2}&{ - 1}&0\end{array}} \right]\)

At row 5, multiply row 3 with 4 and add it to row 5, i.e., \({R_5} \to {R_5} + 3{R_4}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&4&{ - 1}&0&0\\0&0&{ - 4 + 4\left( 1 \right)}&{13 + 0}&{ - 2 + 0}&{ - 1 + 4\left( { - \frac{1}{{18}}} \right)}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&4&{ - 1}&0&0\\0&0&0&{13}&{ - 2}&{ - \frac{{11}}{9}}&0\end{array}} \right]\)

Divide row 4 by 4, i.e., \({R_4} \to \frac{1}{4}{R_4}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&{ - \frac{1}{4}}&0&0\\0&0&0&{13}&{ - 2}&{ - \frac{{11}}{9}}&0\end{array}} \right]\)

At row 5, multiply row 4 with 13 and subtract it from row 5, i.e., \({R_5} \to {R_5} - 13{R_4}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&{ - \frac{1}{4}}&0&0\\0&0&0&{13 - 13\left( 1 \right)}&{ - 2 - 13\left( { - \frac{1}{4}} \right)}&{ - \frac{{11}}{9} - 0}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&{ - \frac{1}{4}}&0&0\\0&0&0&0&{\frac{5}{4}}&{ - \frac{{11}}{9}}&0\end{array}} \right]\)

Multiply row 5 with \(\frac{4}{5}\), i.e., \({R_5} \to \frac{4}{5}{R_5}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&{ - \frac{1}{4}}&0&0\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

For row 4, multiply row 5 with \(\frac{1}{4}\) and add it to row 4, i.e., \({R_4} \to {R_4} + \frac{1}{4}{R_5}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\{0 + 0}&{0 + 0}&{0 + 0}&{1 + 0}&{ - \frac{1}{4} + \frac{1}{4}\left( 1 \right)}&{0 + \frac{1}{4}\left( { - \frac{{44}}{{45}}} \right)}&{0 + 0}\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&{ - 2}&0&0&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&0&{ - \frac{{11}}{{45}}}&{0 + 0}\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

At row 2, multiply row 4 with 2 and add with row 2, i.e., \({R_2} \to {R_2} + 2{R_4}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\{0 + 0}&{1 + 0}&{0 + 0}&{ - 2 + 2\left( 1 \right)}&{0 + 0}&{0 + 2\left( { - \frac{{11}}{{45}}} \right)}&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&0&{ - \frac{{11}}{{45}}}&{0 + 0}\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0&0\\0&1&0&0&0&{ - \frac{{22}}{{45}}}&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&0&{ - \frac{{11}}{{45}}}&{0 + 0}\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

At row 1, multiply row 3 with 3 and add it to row 1, i.e., \({R_1} \to {R_1} + 3{R_3}\).

\(\left[ {\begin{array}{*{20}{c}}{1 + 0}&{0 + 0}&{ - 3 + 3\left( 1 \right)}&{0 + 0}&{0 + 0}&{0 + 3\left( { - \frac{1}{{18}}} \right)}&0\\0&1&0&0&0&{ - \frac{{22}}{{45}}}&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&0&{ - \frac{{11}}{{45}}}&{0 + 0}\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&{ - \frac{1}{6}}&0\\0&1&0&0&0&{ - \frac{{22}}{{45}}}&0\\0&0&1&0&0&{\frac{{ - 1}}{{18}}}&0\\0&0&0&1&0&{ - \frac{{11}}{{45}}}&{0 + 0}\\0&0&0&0&1&{ - \frac{{44}}{{45}}}&0\end{array}} \right]\)

Thegeneral solution for the coefficients of the chemical reaction is:

\({x_1} = \frac{1}{6}{x_6}\), \({x_2} = \frac{{22}}{{45}}{x_6}\), \({x_3} = \frac{1}{{18}}{x_6}\), \({x_4} = \frac{{11}}{{45}}{x_6}\), and \({x_5} = \frac{{44}}{{45}}{x_6}\).

Let \({x_6} = 90\), then

\({x_1} = 15\), \({x_2} = 44\), \({x_3} = 5\), \({x_4} = 22\), and \({x_5} = 88\)

So, the balanced chemical reaction is:

\(15Pb{N_6} + 44CrM{n_2}{O_8} \to 5P{b_3}{O_4} + 22C{r_2}{O_3} + 88Mn{O_2} + 90NO\)