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

Alka-Seltzer contains sodium bicarbonate \(\left( {NaHC{O_3}} \right)\) and citric acid \(\left( {{H_3}{C_6}{H_5}{O_7}} \right)\). When a tablet is dissolved in water, the following reaction produces sodium citrate, water and carbon dioxide (gas):

\(NaHC{O_3} + {H_3}{C_6}{H_5}{O_7} \to N{a_3}{C_6}{H_5}{O_7} + {H_2}O + C{O_2}\)

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 sodium (Na), hydrogen (H), carbon (C), and oxygen (O) for the chemical compounds participating in the chemical reaction.

\({\rm{NaHC}}{{\rm{O}}_3}:\left[ {\begin{array}{*{20}{c}}1\\1\\1\\3\end{array}} \right]\), \({{\rm{H}}_3}{{\rm{C}}_6}{{\rm{H}}_5}{{\rm{O}}_7}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\8\\6\\7\end{array}} \right]\), \({\rm{N}}{{\rm{a}}_3}{{\rm{C}}_6}{{\rm{H}}_5}{{\rm{O}}_7}:\left[ {\begin{array}{*{20}{c}}3\\5\\6\\7\end{array}} \right]\), \({{\rm{H}}_2}{\rm{O:}}\left[ {\begin{array}{*{20}{c}}0\\2\\0\\1\end{array}} \right]\)and \({\rm{C}}{{\rm{O}}_2}{\rm{:}}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\2\end{array}} \right]\)

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

Let the chemical reaction be:

\({x_1} \cdot NaHC{O_3} + {x_2} \cdot {H_3}{C_6}{H_5}{O_7} \to {x_3} \cdot N{a_3}{C_6}{H_5}{O_7} + {x_4} \cdot {H_2}O + C{O_2}\)

The above reaction must satisfy the equation:

\({x_1}\left[ {\begin{array}{*{20}{c}}1\\1\\1\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\8\\6\\7\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}3\\5\\6\\7\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}0\\2\\0\\1\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\2\end{array}} \right]\)

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

The augmented matrix for the equation \({x_1}\left[ {\begin{array}{*{20}{c}}1\\1\\1\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\8\\6\\7\end{array}} \right] = {x_3}\left[ {\begin{array}{*{20}{c}}3\\5\\6\\7\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}0\\2\\0\\1\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\2\end{array}} \right]\) is:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\1&8&{ - 5}&{ - 2}&0&0\\1&6&{ - 6}&0&{ - 1}&0\\3&7&{ - 7}&{ - 1}&{ - 2}&0\end{array}} \right]\).

The row operations do not affect thelinear system.

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

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\1&8&{ - 5}&{ - 2}&0&0\\1&6&{ - 6}&0&{ - 1}&0\\{3 - 1\left( 3 \right)}&{7 - 0\left( 3 \right)}&{ - 7 - 3\left( { - 3} \right)}&{ - 1 - 0\left( 3 \right)}&{ - 2 - 0\left( 3 \right)}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\1&8&{ - 5}&{ - 2}&0&0\\1&6&{ - 6}&0&{ - 1}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

For row 3, subtract row 1 from row 3, i.e., \({R_3} \to {R_3} - {R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\1&8&{ - 5}&{ - 2}&0&0\\{1 - 1}&{6 - 0}&{ - 6 + 3}&{0 - 0}&{ - 1 - 0}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\1&8&{ - 5}&{ - 2}&0&0\\0&6&{ - 3}&0&{ - 1}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\{1 - 1}&{8 - 0}&{ - 5 + 3}&{ - 2 - 0}&{0 - 0}&{0 - 0}\\0&6&{ - 3}&0&{ - 1}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&8&{ - 2}&{ - 2}&0&0\\0&6&{ - 3}&0&{ - 1}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

For row 2, divide row 2 by 2, i.e., \({R_2} \to \frac{1}{2}{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&{\frac{8}{2}}&{\frac{{ - 2}}{2}}&{\frac{{ - 2}}{2}}&0&0\\0&6&{ - 3}&0&{ - 1}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&4&{ - 1}&{ - 1}&0&0\\0&6&{ - 3}&0&{ - 1}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

For row 3, divide row 3 by 3, i.e., \({R_3} \to \frac{1}{3}{R_3}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&4&{ - 1}&{ - 1}&0&0\\0&{\frac{6}{3}}&{\frac{{ - 3}}{3}}&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&4&{ - 1}&{ - 1}&0&0\\0&2&{ - 1}&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

For row 2, apply the row operation \({R_2} \to {R_4} - \left( {{R_3} + {R_2}} \right)\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&{7 - \left( {4 + 2} \right)}&{2 - \left( { - 1 - 1} \right)}&{ - 1 - \left( {0 - 1} \right)}&{ - 2 - \left( {0 - \frac{1}{3}} \right)}&0\\0&2&{ - 1}&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&1&4&0&{ - \frac{5}{3}}&0\\0&2&{ - 1}&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

For row 3, multiply row 2 with 2 and subtract it from row 3.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&1&4&0&{ - \frac{5}{3}}&0\\0&{2 - 1\left( 2 \right)}&{ - 1 - 2\left( 4 \right)}&0&{ - \frac{1}{3} + \frac{{10}}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&1&4&0&{ - \frac{5}{3}}&0\\0&0&{ - 9}&0&3&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

For row 3, divide row 3 by \( - 9\) , i.e., \({R_3} \to - \frac{1}{9}{R_3}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&1&4&0&{ - \frac{5}{3}}&0\\0&0&1&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&{1 - 0}&{4 - 1\left( 4 \right)}&0&{ - \frac{5}{3} + \frac{4}{3}}&0\\0&0&1&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

After the row operation, the matrix will become:

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&1&0&0&{ - \frac{1}{3}}&0\\0&0&1&0&{ - \frac{1}{3}}&0\\0&7&2&{ - 1}&{ - 2}&0\end{array}} \right]\)

Apply row operation \({R_4} \to {R_4} - 7{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0&0\\0&1&0&0&{ - \frac{1}{3}}&0\\0&0&1&0&{ - \frac{1}{3}}&0\\0&{7 - 1\left( 7 \right)}&{2 - 0}&{ - 1 - 0}&{ - 2 + \frac{7}{3}}&0\end{array}} \right]\)

After the row operation, the matrix will become:

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

Apply row operation \({R_4} \to {R_4} - 2{R_3}\).

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

After the row operation, the matrix will become:

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

Apply row operation \({R_4} \to - {R_4}\) and \({R_1} \to {R_1} + 3{R_3}\).

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

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

\({x_1} = {x_5}\), \({x_2} = \frac{1}{3}{x_5}\), \({x_3} = \frac{1}{3}{x_5}\), \({x_4} = {x_5}\)

Let \({x_5} = 3\), then \({x_1} = 3\), \({x_2} = 1\), \({x_3} = 1\), and \({x_4} = 3\)

So, the balanced equation is:

\(3NaHC{O_3} + {H_3}{C_6}{H_5}{O_7} \to N{a_3}{C_6}{H_5}{O_7} + 3{H_2}O + 3C{O_2}\)