Consider the following reaction :

$$ 3 \mathrm{PbCl}_2+2\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \rightarrow \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2+6 \mathrm{NH}_4 \mathrm{Cl} $$

If $72 ~\mathrm{mmol}$ of $\mathrm{PbCl}_2$ is mixed with $50 ~\mathrm{mmol}$ of $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$, then the amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed is ________ mmol (nearest integer).

<p>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 $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$.</p> <p>The reaction is:</p> $$ 3 \mathrm{PbCl}_2 + 2\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \rightarrow \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2 + 6 \mathrm{NH}_4 \mathrm{Cl} $$ <p>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.</p> <p>The stoichiometric calculations are as follows:</p> <p>For $\mathrm{PbCl}_2$:</p> $$ \text{Moles of }\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2\text{ formed from } \mathrm{PbCl}_2 = \frac{72 \text{ mmol of } \mathrm{PbCl}_2}{3 \text{ mmol of } \mathrm{PbCl}_2\text{/mmol of } \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2} = 24 \text{ mmol} $$ <p>For $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$:</p> $$ \text{Moles of }\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2\text{ formed from } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 = \frac{50 \text{ mmol of } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4}{2 \text{ mmol of } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\text{/mmol of } \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2} = 25 \text{ mmol} $$ <p>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.</p> <p>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.</p> <p>Therefore, the amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed is 24 mmol (as a nearest integer).</p>