1. Write the balanced reaction

$3\,\mathrm{PbCl_2} + 2\,(\mathrm{NH_4})_3\mathrm{PO_4} \rightarrow \mathrm{Pb_3(PO_4)_2} + 6\,\mathrm{NH_4Cl}$

2. Given amounts

• $\mathrm{PbCl_2} = 72\ \mathrm{mmol}$
• $(\mathrm{NH_4})_3\mathrm{PO_4} = 50\ \mathrm{mmol}$

3. Use stoichiometric ratios to find the limiting reagent

From the equation:

$3\ \mathrm{mol\ PbCl_2} \text{ react with } 2\ \mathrm{mol\ }(\mathrm{NH_4})_3\mathrm{PO_4}$

For $72\ \mathrm{mmol}$ of $\mathrm{PbCl_2}$, required $(\mathrm{NH_4})_3\mathrm{PO_4}$ is:

$72 \times \frac{2}{3} = 48\ \mathrm{mmol}$

Available $(\mathrm{NH_4})_3\mathrm{PO_4}$ is $50\ \mathrm{mmol}$, so it is in excess.

Hence, $\mathrm{PbCl_2}$ is the limiting reagent.

4. Calculate product formed

From the balanced equation:

$3\ \mathrm{mol\ PbCl_2} \rightarrow 1\ \mathrm{mol\ Pb_3(PO_4)_2}$

So, from $72\ \mathrm{mmol}$ of $\mathrm{PbCl_2}$:

$\mathrm{Pb_3(PO_4)_2} = 72 \times \frac{1}{3} = 24\ \mathrm{mmol}$

5. Nearest integer

$24\ \mathrm{mmol}$

So, the amount of $\mathrm{Pb_3(PO_4)_2}$ formed is:

$\boxed{24}$

6. Comparison with stored answer

Stored correct answer = $24$

Our derived answer = $24$

They match.