The reaction is given as,

$3{\mathrm{PbCl}}_2+2{\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4\to {\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2+6{\mathrm{NH}}_4\mathrm{Cl}$

The products formed in the above reaction is ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2 \text{and} {\mathrm{NH}}_4\mathrm{Cl}$  .

According to the above stoichiometric equation, three mole of each lead chloride and two moles of ${\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4$ is required to make one mole of ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2\text{ and} 6\text{ moles of} {\mathrm{NH}}_4\mathrm{Cl}$   

Here the Limiting Reagent is ${\mathrm{PbCl}}_2$

$\mathrm{mmol}$ of ${\mathrm{Pb}}_3({\mathrm{PO}}_4{)}_2$ formed

$$\begin{gathered} =\frac{\mathrm{mmol}\text{ of }{\mathrm{PbCl}}_2\text{ reacted }}{3} \\ =\frac{72}{3} \end{gathered}$$

$=24 \mathrm{m mol}$