$3CaC{{l}_{2}}+2N{{a}_{3}}P{{O}_{4}}\to C{{a}_{3}}{{(P{{O}_{4}})}_{2}}+6NaCl$ $?mLCaC{{l}_{2}}(aq)=1/312gN{{a}_{3}}P{{O}_{4}}\times \frac{1molN{{a}_{3}}P{{O}_{4}}}{164gN{{a}_{3}}P{{O}_{4}}}\times \frac{3molCaC{{l}_{2}}}{2molN{{a}_{3}}P{{O}_{4}}}\times \frac{1LCaC{{l}_{2}}(aq)}{0/06molCaC{{l}_{2}}}\times \frac{1000mL}{1L}=200mL$ $?molIon=1/312gN{{a}_{3}}P{{O}_{4}}\times \frac{1molN{{a}_{3}}P{{O}_{4}}}{164gN{{a}_{3}}P{{O}_{4}}}\times \frac{6molNaCl}{2molN{{a}_{3}}P{{O}_{4}}}\times \frac{2molIon}{1molNaCl}=0/048molIon$