Upon mixing $50.0mL$ of $0.1M$ lead nitrate solution with $50.0mL$ of $0.05M$ chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentration of the species left behind in the final solution. Which is the limiting reagent?

Solution: The reaction involved is
$3Pb{\left(N{O}_3\right)}_2+C{r}_2{\left(S{O}_4\right)}_3\to 3PbS{O}_4\left(s\right)\downarrow +2Cr{\left(N{O}_3\right)}_3$
millimol of $Pb{\left(N{O}_3\right)}_2$ taken $=45\times 0.25=11.25$
millimol of $C{r}_2{\left(S{O}_4\right)}_3$ taken $=2.5$
Here, chromic sulphate is the limiting reagent, it will determine the amount of product.
$\because 1$ mole $C{r}_2{\left(S{O}_4\right)}_3$ will produces $3$ mole $PbS{O}_4$.
$\therefore 2.5$ millimol $C{r}_2{\left(S{O}_4\right)}_3$ will produce $7.5$ millimol $PbS{O}_4$
Hence, mole of $PbS{O}_4$ precipitate formed $=7.5\times {10}^{-3}$
Also, millimol of $Pb{\left(N{O}_3\right)}_2$ remaining unreacted
$11.25-7.50=3.75$
$\Rightarrow$ Molarity of $Pb{\left(N{O}_3\right)}_2$ in final solution
$=\frac{\text{millimol of }Pb{\left(N{O}_3\right)}_2}{\text{Total volume}}=\frac{3.75}{70}=0.054M$
Also, millimol of $Cr{\left(N{O}_3\right)}_2$ formed
$=2\times$ millimol of $C{r}_2{\left(S{O}_4\right)}_3$ reacted
$\Rightarrow$ Molarity of $Cr{\left(N{O}_3\right)}_2=\frac{5}{70}=0.071M$