JEE Main 2024 — Some Basic Concepts Of Chemistry Question with Solution

To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced equation shows that 3 moles of ${\mathrm{PbCl}}_2$ react with 2 moles of ${\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4$ to produce 1 mole of ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$.

The reaction is:

The molar ratio of ${\mathrm{PbCl}}_2$ to ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$ is 3:1, and the molar ratio of ${\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4$ to ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$ is 2:1. We need to determine which reactant is the limiting reagent because it will dictate the amount of ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$ produced.

The stoichiometric calculations are as follows:

For ${\mathrm{PbCl}}_2$:

For ${\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4$:

Now we can identify the limiting reagent by comparing the two amounts of ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$ that could be produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts the reaction.

Since the ${\mathrm{PbCl}}_2$ can produce only 24 mmol of ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$ versus the 25 mmol that could be produced by ${\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4$, ${\mathrm{PbCl}}_2$ is the limiting reagent.

Therefore, the amount of ${\mathrm{Pb}}_3{\left({\mathrm{PO}}_4\right)}_2$ formed is 24 mmol (as a nearest integer).