Balance the chemical equations in Exercise 5-10 using the vector equation approach discussed in this section.\(\left( {KMn{O_4}} \right)\)

The following reaction between potassium permanganate and manganese sulfate in water produces manganese dioxide, potassium sulfate and sulfuric acid:

\(KMn{O_4} + MnS{O_4} + {H_2}O \to Mn{O_2} + {K_2}S{O_4} + {H_2}S{O_4}\)

[For each compound, construct a vector that lists the numbers of atoms of potassium (K), manganese, oxygen, sulfur and hydrogen.]

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

The following vectors represent the number of atoms of potassium (K), manganese (Mn), oxygen (O), sulfur (S), and hydrogen (H) for the chemical compounds participating in the chemical reaction.

\({\rm{KMn}}{{\rm{O}}_4}:\left[ {\begin{array}{*{20}{c}}1\\1\\4\\0\\0\end{array}} \right]\), \({\rm{MnS}}{{\rm{O}}_4}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\1\\4\\1\\0\end{array}} \right]\), \({{\rm{H}}_2}{\rm{O}}:\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\\2\end{array}} \right]\), \({\rm{Mn}}{{\rm{O}}_2}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\1\\2\\0\\0\end{array}} \right]\), \({{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4}{\rm{:}}\left[ {\begin{array}{*{20}{c}}2\\0\\4\\1\\0\end{array}} \right]\)and \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\0\\4\\1\\2\end{array}} \right]\)

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

Let the chemical reaction be:

\({x_1} \cdot {\rm{KMn}}{{\rm{O}}_4} + {x_2} \cdot {\rm{MnS}}{{\rm{O}}_4} + {x_3} \cdot {{\rm{H}}_2}{\rm{O}} \to {x_4} \cdot {\rm{Mn}}{{\rm{O}}_2} + {x_5} \cdot {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + {x_6} \cdot {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\)

The above reaction must satisfy the equation:

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

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

The augmented matrixfor the equation \({x_1}\left[ {\begin{array}{*{20}{c}}1\\1\\4\\0\\0\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\1\\4\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\\2\end{array}} \right] = {x_4}\left[ {\begin{array}{*{20}{c}}0\\1\\2\\0\\0\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}2\\0\\4\\1\\0\end{array}} \right] + {x_6}\left[ {\begin{array}{*{20}{c}}0\\0\\4\\1\\2\end{array}} \right]\) is:

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

The row operations do not affect thelinear system.

For row 3, multiply row 1 with 4 and subtract it from row 4, i.e., \({R_4} \to {R_4} - 4{R_1}\).

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

After row operation, the matrix will become:

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

For row 2, subtract row 1 from row 2, i.e., \({R_2} \to {R_2} - {R_1}\).

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

After the row operation, the matrix will become:

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

For row 3, multiply row 2 with 4 and subtract it from row 3, i.e., \({R_3} \to {R_3} - 4{R_2}\).

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

After row operations, the matrix will become:

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

For row 4, subtract row 2 from row 4, i.e., \({R_4} \to {R_4} - {R_2}\).

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

After the row operation, the matrix will become:

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

For row 5, multiply row 3 with 2 and subtract it from row 5, i.e., \({R_5} \to {R_5} - 2{R_3}\).

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

After the row operation, the matrix will become:

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

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

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

After the row operation, the matrix will become:

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

Multiply row 5 with \( - \frac{1}{4}\) , i.e., \({R_5} \to - \frac{1}{4}{R_5}\).

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

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

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

After the row operation, the matrix will become:

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

For row 3, multiply row 5 with 4 and add it to row 3, i.e., \({R_3} \to {R_3} + 4{R_5}\).

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

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&{ - 2}&0&0\\0&1&0&{ - 1}&2&0&0\\0&0&1&2&0&{ - 6}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&{ - 2}&0&0\\{0 - 0}&{1 - 0}&{0 - 0}&{ - 1 - 0}&{2 - 2\left( 1 \right)}&{0 - 2\left( { - \frac{1}{2}} \right)}&0\\0&0&1&2&0&{ - 6}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&{ - 2}&0&0\\0&1&0&{ - 1}&0&1&0\\0&0&1&2&0&{ - 6}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

For row 1, multiply row 5 with 2 and add it to row 1, i.e., \({R_1} \to {R_1} + 2{R_5}\).

\(\left[ {\begin{array}{*{20}{c}}{1 + 0}&{0 + 0}&{0 + 0}&{0 + 0}&{ - 2 + 2\left( 1 \right)}&{0 + 2\left( { - \frac{1}{2}} \right)}&0\\0&1&0&{ - 1}&0&1&0\\0&0&1&2&0&{ - 6}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&{ - 1}&0\\0&1&0&{ - 1}&0&1&0\\0&0&1&2&0&{ - 6}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

For row 3, multiply row 4 with 2 and subtract it from row 3, i.e., \({R_3} \to {R_3} - 2{R_4}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&{ - 1}&0\\0&1&0&{ - 1}&0&1&0\\{0 - 0}&{0 - 0}&{1 - 0}&{2 - 2\left( 1 \right)}&{0 - 0}&{ - 6 - 2\left( { - \frac{5}{2}} \right)}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

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

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

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0&0&{ - 1}&0\\{0 + 0}&{1 + 0}&{0 + 0}&{ - 1 + 1}&{0 + 0}&{1 - \frac{5}{2}}&0\\0&0&1&0&0&{ - 1}&0\\0&0&0&1&0&{ - \frac{5}{2}}&0\\0&0&0&0&1&{ - \frac{1}{2}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

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

The general solution for the coefficients of the chemical reaction is:

\({x_1} = {x_6}\), \({x_2} = \frac{3}{2}{x_6}\), \({x_3} = {x_6}\), \({x_4} = \frac{5}{2}{x_6}\) and \({x_5} = \frac{1}{2}{x_6}\).

Let \({x_6} = 2\), then

\({x_1} = 2\), \({x_2} = 3\), \({x_3} = 2\), \({x_4} = 5\) and \({x_5} = 1\)

So, the balanced chemical reaction is:

\(2{\rm{KMn}}{{\rm{O}}_4} + 3{\rm{MnS}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{O}} \to 5{\rm{Mn}}{{\rm{O}}_2} + {{\rm{K}}_2}{\rm{S}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\)