Question 2.4.7: Balance the chemical equation (NH4)3PO4 + Pb(NO3)4 → Pb3(PO4...

We want to find constants x_{1}, x_{2}, x_{3}, x_{4} such that

x_{1}(NH_{4})_{3}PO_{4} + x_{2}Pb(NO_{3})_{4} \rightarrow x_{3}Pb_{3}(PO_{4})_{4} + x_{4}NH_{4}NO_{3}

is balanced. Define vectors in \mathbb{R} ^5 by

We get the vector equation

x_{1}\left [ \begin{matrix} 3 \\ 12 \\ 1 \\ 4 \\ 0 \end{matrix} \right ] + x_{2}\left [ \begin{matrix} 4 \\ 0 \\ 0 \\ 12 \\ 1 \end{matrix} \right ] = x_{3}\left [ \begin{matrix} 0 \\ 0 \\ 4 \\ 16 \\ 3 \end{matrix} \right ] + x_{4}\left [ \begin{matrix} 2 \\ 4 \\ 0 \\ 3 \\ 0 \end{matrix} \right]

Rearranging gives the homogeneous system

    3x_{1} + 4x_{2} – 2x_{4} = 0

               12x_{1} – 4x_{4} = 0

                       x_{1} – 4x_{3} = 0

4x_{1} + 12x_{2} – 16x_{3} – 3x_{4} = 0

                       x_{2} – 3x_{3} = 0

Row reducing the corresponding coefficient matrix gives

\left [ \begin{matrix} 3 & 4 & 0 & -2 \\ 12 & 0 & 0 & -4 \\ 1 & 0 & -4 & 0 \\ 4 & 12 & -16 & -3 \\ 0 & 1 & -3 & 0 \end{matrix} \right ] \thicksim \left [ \begin{matrix} 1 & 0 & 0 & -1/3 \\ 0 & 1 & 0 & -1/4 \\ 0 & 0 & 1 & -1/12 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right]

We find that a vector equation for the solution space is

\left [ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix} \right ] = t\left [ \begin{matrix} 1/3 \\ 1/4 \\ 1/12 \\ 1 \end{matrix} \right ] , \ \ \ \ t \in \mathbb{R}

To get the smallest positive integer values, we take t = 12. This gives x_{1} = 4, x_{2} = 3, x_{3} = 1 and x_{4} = 12. Thus, a balanced chemical equation is

4(NH_{4})_{3}PO_{4} + 3Pb(NO_{3})_{4} \rightarrow Pb_{3}(PO_{4})_{4} + 12NH_{4}NO_{3}