Question

Consider the following reaction:\[3\text{PbCl}_2 + 2(\text{NH}_4)_3\text{PO}_4 \rightarrow \text{Pb}_3(\text{PO}_4)_2 + 6\text{NH}_4\text{Cl}\]If 72 mmol of PbCl\(_2\) is mixed with 50 mmol of (\(\text{NH}_4\))\(_3\)PO\(_4\), then the amount of \(\text{Pb}_3(\text{PO}_4)_2\) formed is ______ mmol (nearest integer).

Answer 1

Correct Answer: 24

Limiting reagent is PbCl₂.

Amount of Pb₃(PO₄)₂ formed:

\( \text{mmol of Pb}_3(\text{PO}_4)_2 = \frac{\text{mmol of PbCl}_2 \text{ reacted}}{3} = 24 \text{ mmol} \)