Upon mixing 50.0 mL of 0.1 M lead nitrate solution with 50 mL of 0.05 M chromic sulphate solution, precipitation of lead sulphate solution takes place. How many moles of lead sulphate are formed and what is the molar concentration of chromic sulphate left in the solution?

$\begin{array}{l}3 \mathrm{mmol} 1 \mathrm{mmol} 3 \mathrm{mmol} 2\mathrm{mmol}\\ 3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\to 3{\mathrm{PbSO}}_4\downarrow +2\mathrm{Cr}{\left({\mathrm{NO}}_3\right)}_3\\ \begin{array}{l}50\times 0.1\end{array} 50 \times 0.05 \\ =5 mmol =2.5 mmol\\ \text{ First find the limiting reagent. }\\ 3 \mathrm{mmol}\text{ of }\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\Rightarrow 1 \mathrm{mmol}\text{ of }{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\\ 5 \mathrm{mmol}\text{ of }\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\Rightarrow \frac{1}{3}\times 5\\ \Rightarrow \frac{5}{3}=1.66\mathrm{mmol}\\ \text{ So }\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\text{ is the limiting reagent. }\\ \mathbf{i}\mathbf{.} 3\mathrm{mmol}\text{ of }\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\Rightarrow 3 \mathrm{mmol}\text{ of }{\mathrm{PbSO}}_4\\ 5 \mathrm{mmol}\text{ of }\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\Rightarrow 5 \mathrm{mmol}\\ \Rightarrow \frac{5}{1000}\mathrm{mol}\Rightarrow 0.005 \mathrm{mol}\\ \text{ii. Species left in the solution are }{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\text{ and }\mathrm{Cr}{\left({\mathrm{NO}}_3\right)}_3\text{. }\\ \text{ To calculate the concentration of }{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\text{ : }\\ \text{ Initial }\mathrm{mmol}=2.5\\ \text{ Reacted }\mathrm{mmol}=1.65\\ \text{ Left mmoles }=2.5-1.66=0.84\mathrm{mmol}\\ \text{ Total Volume }=50+50=100\mathrm{mL}\\ \text{ Concentration }=\frac{0.84}{100}=0.0084\mathrm{M}\end{array}$

iii. To calculate the concentration of $\mathrm{Cr}{\left({\mathrm{NO}}_3\right)}_3$

3 m mol of Pb(NO₃)₂ $\Rightarrow \frac{2}{3}\times 5$

$\Rightarrow \frac{10}{3}=3.33 \mathrm{mmol}$

Concentration = $\frac{3.33}{100}=0.033\mathrm{M}$