If $0.50\mathrm{mol}$ of ${\mathrm{CaCl}}_2$ is mixed with $0.20 mol$ of ${\mathrm{Na}}_3{\mathrm{PO}}_4$ the maximum number of moles of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ which can be formed, is

$3{\mathrm{CaCl}}_2+2{\mathrm{Na}}_3{\mathrm{PO}}_4\to {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+6\mathrm{NaCl}$

$\therefore \text{ 2Moles of }{\mathrm{Na}}_3{\mathrm{PO}}_4=3\text{ mole of }{\mathrm{CaCl}}_2=1\text{ mole }$${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

$\therefore 0.2\text{ mole of }{\mathrm{Na}}_3{\mathrm{PO}}_4=0.3\text{ mole of }{\mathrm{CaCl}}_2=0.1$$\text{ mole of }{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\text{. }$