Consider the following reaction: $$3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl$$. If $$72$$ mmol $$PbCl_2$$ is mixed with $$50$$ mmol of $$(NH_4)_3PO_4$$, then amount of $$Pb_3(PO_4)_2$$ formed in mmol is (nearest integer):

We are asked to find the number of millimoles of $$Pb_3(PO_4)_2$$ formed when 72 mmol of $$PbCl_2$$ reacts with 50 mmol of $$(NH_4)_3PO_4$$.

The reaction is represented by the balanced chemical equation $$ 3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl $$. From this equation, the stoichiometric ratio is 3 mol of $$PbCl_2$$ to 2 mol of $$(NH_4)_3PO_4$$ to 1 mol of $$Pb_3(PO_4)_2$$.

Comparing the available amounts, 72 mmol of $$PbCl_2$$ would require $$\frac{2}{3} \times 72 = 48$$ mmol of $$(NH_4)_3PO_4$$, and since 50 mmol of $$(NH_4)_3PO_4$$ is available, there is more than enough of the phosphate. Conversely, 50 mmol of $$(NH_4)_3PO_4$$ would require $$\frac{3}{2} \times 50 = 75$$ mmol of $$PbCl_2$$, but only 72 mmol of $$PbCl_2$$ are present. Therefore, $$PbCl_2$$ is the limiting reagent.

According to the stoichiometry, 3 mmol of $$PbCl_2$$ yields 1 mmol of $$Pb_3(PO_4)_2$$, so the amount of product formed is $$ \text{mmol of } Pb_3(PO_4)_2 = \frac{72}{3} = 24 \text{ mmol} $$.

The answer is 24 mmol.