QJEEM24Ch46c9: JEE Main 2024 Chemistry - Class 11 JEEM24Ch46c9 Solution: Consider the following reaction : If of is mixed with of , then …

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).