\( - {n_{C{a_3}{{(P{O_4})}_2}}}{\rm{\;}} = {\rm{\;}}0,1{\rm{\;}}(mol){\rm{\;}} \Rightarrow {\rm{\;}}{n_{P{O_4}^{3 - }}}{\rm{\;}} = {\rm{\;}}0,2{\rm{\;}}(mol){\rm{\;}} \Rightarrow {\rm{\;}}{C_M}{\rm{\;}} = {\rm{\;}}0,2{\rm{\;}}M\)

- Áp dụng BTĐT và BTKL ta có

\(\begin{array}{l} \left\{ \begin{array}{l} {n_{{K^ + }}} + {n_{N{a^ + }}} = 3{n_{P{O_4}^{3 - }}}\\ {m_{rankhan}} = {m_{{K^ + }}} + {m_{N{a^ + }}} + {m_{P{O_4}^{3 - }}} \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} 1.x + 1.y\; = \;3.0,2\\ 39x + 23y + 95.0,2\; = \;37,6 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} x\; = \;0,3\; = \;{n_{{K^ + }}} \Rightarrow \;{C_M} = \;0,3\;M\\ y\; = \;0,3\; = \;{n_{N{a^ + }}}\; \Rightarrow \;{C_M} = \;0,3\;M \end{array} \right. \end{array}\)