Question 2.4.6: Balance the chemical equation H3PO4 + Mg(OH)2 → Mg3(PO4)2 + ...

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

$x_{1}H_{3}PO_{4} \ + \ x_{2}Mg(OH)_{2} \rightarrow x_{3}Mg_{3}(PO_{4})_{2} \ + x_{4}H_{2}O$

is balanced. To turn this into a vector equation, we represent the molecules in the equation with the vectors in $\mathbb{R} ^4$ :

We get

$x_{1}\left [ \begin{matrix} 3 \\ 1 \\ 4 \\ 0 \end{matrix} \right ] + x_{2}\left [ \begin{matrix} 2 \\ 0 \\ 2 \\ 1 \end{matrix} \right ] = x_{3}\left [ \begin{matrix} 0 \\ 2 \\ 8 \\ 3 \end{matrix} \right ] + x_{4} \left [ \begin{matrix} 2 \\ 0 \\ 1 \\ 0 \end{matrix} \right ]$

Moving all the terms to the left side and performing the linear combination of vectors, we get the homogeneous system

$\ \ \ \ \ \ \ \ \ 3x_{1} + 2x_{2} – 2x_{4} = 0\\$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{1} – 2x_{3} = 0 \\$

$\ \ \ \ 4x_{1} + 2x_{2} – 8x_{3} – x_{4} = 0 \\$

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

Row reducing the corresponding coefficient matrix gives

$\left [ \begin{matrix} 3 & 2 & 0 & -2 \\ 1 & 0 & -2 & 0 \\ 4 & 2 & -8 & -1 \\ 0 & 1 & -3 & 0 \end{matrix} \right ] \thicksim \left [ \begin{matrix} 1 & 0 & 0 & -1/3 \\ 0 & 1 & 0 & -1/2 \\ 0 & 0 & 1 & -1/6 \\ 0 & 0 & 0 & 0 \end{matrix} \right ]$

We find that a vector equation for the solution space is

$\left [ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix} \right ] = t\left [ \begin{matrix} 1/3 \\ 1/2 \\ 1/6 \\ 1 \end{matrix} \right ] , \ \ \ \ \ \ t\in \mathbb{R}$

To get the smallest positive integer values, we take $t = 6$. This gives $x_{1} = 2, x_{2} = 3, x_{3} = 1$ and $x_{4} = 6$.Thus, a balanced chemical equation is

$2H_{3}PO_{4} \ + \ 3Mg(OH)_{2} \rightarrow Mg_{3}(PO_{4})_{2} \ + 6H_{2}O$