Balance the chemical equations in Exercise 5-10 using the vector equation approach discussed in this section.

When solution of sodium phosphate and barium nitrate are mixed, the result is barium phosphate (as a precipitate) and sodium nitrate. The unbalanced equation is

\(N{a_3}P{O_4} + Ba{\left( {N{O_3}} \right)_2} \to B{a_3}{\left( {P{O_4}} \right)_2} + NaN{O_3}\)

[For each compound, construct a vector that lists the numbers of atoms of sodium (Na), Phosphorus, oxygen, barium nitrate correspond to (0, 0, 6, 1, 2).]

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

The following vectors represent the number of atoms of sodium (Na), phosphorus (P), oxygen (O), barium (B), and nitrogen (N).

\({\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4}:\left[ {\begin{array}{*{20}{c}}3\\1\\4\\0\\0\end{array}} \right]\), \({\rm{Ba}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\0\\6\\1\\2\end{array}} \right]\), \({\rm{B}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2}:\left[ {\begin{array}{*{20}{c}}0\\2\\8\\3\\0\end{array}} \right]\) and \({\rm{NaN}}{{\rm{O}}_3}{\rm{:}}\left[ {\begin{array}{*{20}{c}}1\\0\\3\\0\\1\end{array}} \right]\)

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

Let the chemical reaction be:

\({x_1} \cdot {\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} + {x_2} \cdot {\rm{Ba}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2} \to {x_3} \cdot {\rm{B}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} + {x_4} \cdot {\rm{NaN}}{{\rm{O}}_3}\)

The above reaction must satisfy the equation:

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

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

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

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

The row operations do not affect the linear system.

Exchange row 1 and row 2 in the augmented matrix.

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

For row 2, multiply row 1 with 3 and subtract it from row 1, i.e., \({R_2} \to {R_2} - 3{R_1}\).

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

For row 3, multiply row 1 with 4 and subtract it from row 1, i.e.,\({R_3} \to {R_3} - 4{R_1}\).

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

So, the augmented matrix after the row operation will become:

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

Exchange row 3 and row 4, i.e., \({R_2} \leftrightarrow {R_4}\).

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

For row 3, multiply row 2 with 6 and subtract it from row 3, i.e., \({R_3} \to {R_3} - 6{R_2}\).

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

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0&0\\0&1&{ - 3}&0&0\\0&0&{18}&{ - 3}&0\\0&0&6&{ - 1}&0\\0&2&0&{ - 1}&0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0&0\\0&1&{ - 3}&0&0\\0&0&{18}&{ - 3}&0\\0&0&6&{ - 1}&0\\0&{2 - 1\left( 2 \right)}&{0 - 2\left( { - 3} \right)}&{ - 1 - 0\left( 2 \right)}&0\end{array}} \right]\)

After the row operation, the matrix will become.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0&0\\0&1&{ - 3}&0&0\\0&0&{18}&{ - 3}&0\\0&0&6&{ - 1}&0\\0&0&6&{ - 1}&0\end{array}} \right]\)

For row 5, subtract row 4 from row 5, i.e., \({R_4} \to {R_5} - {R_4}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0&0\\0&1&{ - 3}&0&0\\0&0&{18}&{ - 3}&0\\0&0&6&{ - 1}&0\\0&0&{6 - 6}&{ - 1 - \left( { - 1} \right)}&0\end{array}} \right]\)

After the row operations, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0&0\\0&1&{ - 3}&0&0\\0&0&{18}&{ - 3}&0\\0&0&6&{ - 1}&0\\0&0&0&0&0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0&0\\0&1&{ - 3}&0&0\\0&0&{18}&{ - 3}&0\\0&0&{6\left( 3 \right) - 18}&{ - 1\left( 3 \right) + 3}&0\\0&0&0&0&0\end{array}} \right]\)

After the row operation, the matrix will become:

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

Divide row 3 with 18, i.e., \({R_3} \to \frac{1}{{18}}{R_3}\).

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

After the row operation, the matrix will become:

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

For row 2, multiply row 3 with 3 and add it to row 2, i.e., \({R_2} \to {R_2} + 3{R_3}\).

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

For row 1, multiply row 3 with 2 and add it to row 1, i.e., \({R_1} \to {R_1} + 2{R_3}\).

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

After the row operation, the matrix will become:

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

From the matrix \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - \frac{1}{3}}&0\\0&1&0&{ - \frac{1}{2}}&0\\0&0&1&{\frac{{ - 1}}{6}}&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right]\), thegeneral solution can be written as:

\({x_1} = \frac{1}{3}{x_4}\), \({x_2} = \frac{1}{2}{x_4}\) and \({x_3} = \frac{1}{6}{x_4}\).

Let \({x_4} = 6\), then \({x_1} = 2\), \({x_2} = 3\), and \({x_3} = 1\)

So, the balanced reaction is:

\(2{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} + 3{\rm{Ba}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2} \to {\rm{B}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} + 6{\rm{NaN}}{{\rm{O}}_3}\)