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 mmol of $\mathrm{PbCl}_2$ is mixed with 50 mmol of $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$, then amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed is $\_\_\_\_$ mmol. (nearest integer)
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Solution
Sol. Limiting Reagent is $\mathrm{PbCl}_2 \mathrm{mmol}$ of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed $=\frac{\mathrm{mmol} \text { of } \mathrm{PbCl}_2 \text { reacted }}{3}=24 \mathrm{mmol}$
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