Consider the following reaction: $3\mathrm{PbCl}_{2}+2(\mathrm{NH}_{4})_{3}\mathrm{PO}_{4} \rightarrow \mathrm{~Pb}_{3}(\mathrm{PO}_{4})_{2}+6\mathrm{NH}_{4}\mathrm{Cl}$ If 72 mmol of $\mathrm{PbCl}_{2}$ is mixed with 50 mmol of $(\mathrm{NH}_{4})_{3}\mathrm{PO}_{4}$, then amount of $\mathrm{Pb}_{3}(\mathrm{PO}_{4})_{2}$ formed is ____ mmol. (nearest integer)

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Some Basic Concepts of Chemistry

Consider the following reaction:

3 PbCl_{2} + 2 (NH_{4})_{3} PO_{4} \to Pb_{3} (PO_{4})_{2} + 6 NH_{4} Cl

If 72 mmol of

PbCl_{2}

is mixed with 50 mmol of

(NH_{4})_{3} PO_{4}

, then amount of

Pb_{3} (PO_{4})_{2}

formed is ____ mmol. (nearest integer)

Answer: 24

3 PbCl_{2} + 2 (NH_{4})_{3} PO_{4} \to Pb_{3} (PO_{4})_{2} + 6 NH_{4} Cl

72 mmol PbCl

_{2}

would need

\frac{2}{3} \times 72 = 48

mmol of

(NH_{4})_{3} PO_{4}

, but only 50 mmol is available (barely more than enough), so PbCl

_{2}

is essentially the limiting reagent (using up 48 of the 50 mmol available of the phosphate). From the stoichiometry, 3 mol PbCl

_{2}

gives 1 mol Pb

_{3}

(PO

_{4}

)

_{2}

n (Pb_{3} (PO_{4})_{2}) = \frac{72}{3} = 24 mmol

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer