JEE Main 2024 — Chemistry Physical Chemistry

2024integermedium

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 $72\mathrm{mmol}{\mathrm{PbCl}}_{2}$ is mixed with $50\mathrm{mmol}$ of ${({\mathrm{NH}}_{4})}_{3}{\mathrm{PO}}_{4},$ then amount of ${\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}$ formed in $\mathrm{mmol}$ . (nearest integer)

Official previous-year question

Held on 1 Feb 2024 · Verified 6 Jul 2026.

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Solution

The reaction is given as,

$3{\mathrm{PbCl}}_{2}+2{({\mathrm{NH}}_{4})}_{3}{\mathrm{PO}}_{4}\rightarrow {\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}+6{\mathrm{NH}}_{4}\mathrm{Cl}$

The products formed in the above reaction is ${\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}\text{and}{\mathrm{NH}}_{4}\mathrm{Cl}$ .

According to the above stoichiometric equation, three mole of each lead chloride and two moles of ${({\mathrm{NH}}_{4})}_{3}{\mathrm{PO}}_{4}$ is required to make one mole of ${\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}\text{ and}6\text{ moles of}{\mathrm{NH}}_{4}\mathrm{Cl}$

Here the Limiting Reagent is ${\mathrm{PbCl}}_{2}$

$\mathrm{mmol}$ of ${\mathrm{Pb}}_{3}({\mathrm{PO}}_{4}{)}_{2}$ formed

$=\frac{\mathrm{mmol}\text{ of }{\mathrm{PbCl}}_{2}\text{ reacted }}{3} =\frac{72}{3}$

$=24m mol$

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JEE Main 2024 — Chemistry Physical Chemistry

2024integermedium

Consider the following reaction:

$3{\mathrm{PbCl}}_{2}+2{({\mathrm{NH}}_{4})}_{3}{\mathrm{PO}}_{4}\rightarrow {\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}+6{\mathrm{NH}}_{4}\mathrm{Cl}$

Official previous-year question

Held on 1 Feb 2024 · Verified 6 Jul 2026.

Did you get this right?

Solution

The reaction is given as,

$3{\mathrm{PbCl}}_{2}+2{({\mathrm{NH}}_{4})}_{3}{\mathrm{PO}}_{4}\rightarrow {\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}+6{\mathrm{NH}}_{4}\mathrm{Cl}$

The products formed in the above reaction is ${\mathrm{Pb}}_{3}{({\mathrm{PO}}_{4})}_{2}\text{and}{\mathrm{NH}}_{4}\mathrm{Cl}$ .

Here the Limiting Reagent is ${\mathrm{PbCl}}_{2}$

$\mathrm{mmol}$ of ${\mathrm{Pb}}_{3}({\mathrm{PO}}_{4}{)}_{2}$ formed

$=\frac{\mathrm{mmol}\text{ of }{\mathrm{PbCl}}_{2}\text{ reacted }}{3} =\frac{72}{3}$

$=24m mol$

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2024 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2024 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry