JEE Main 2024 — 1 February, Shift 1 — Question 81

Balanced reaction:

$$\mathrm{3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl}$$

From stoichiometry, we have

$$3\,\mathrm{mol\ PbCl_2} \rightarrow 1\,\mathrm{mol\ Pb_3(PO_4)_2}$$ $$2\,\mathrm{mol\ (NH_4)_3PO_4} \rightarrow 1\,\mathrm{mol\ Pb_3(PO_4)_2}$$

For $72\,\mathrm{mmol}$ $\mathrm{PbCl_2}$, required $\mathrm{(NH_4)_3PO_4}$ $= \frac{2}{3} \times 72 = 48\,\mathrm{mmol}$

Since, $50\,\mathrm{mmol}$ $\mathrm{(NH_4)_3PO_4}$ is available, $\mathrm{PbCl_2}$ is the limiting reagent.

Amount of $\mathrm{Pb_3(PO_4)_2}$ formed $= \frac{72}{3} = 24\,\mathrm{mmol}$

Thus, the amount formed is $24\,\mathrm{mmol}$.