Question

Upon mixing $50$ mL of $0.1M$ lead nitrate solution with $50$ mL of $0.05M$ chromic sulphate solution, precipitation of lead sulphate solution takes place. $\frac{x}{1000}$ moles of lead sulphate are formed. The value of $x$ is :

Answer 1

$\underset{\underset{\underset{=5mmol}{50\times 0.1}}{3Pb\left(N{O}_3{)}_2\right)}}{3mmol}+\underset{C{r}_2)S{O}_4{)}_3}{1mmol}\to \underset{\underset{\underset{=2.5mmol}{50\times 0.05}}{3PbS{O}_4)}}{3mmol}\downarrow +\underset{2mmol}{2Cr(N{O}_3{)}_2}$
First find the limiting reagent.
3 mmol of $Pb(N{O}_3{)}_2\Rightarrow$ 1 mmol of $C{r}_2(S{O}_4{)}_3$
5 mmol of $Pb(N{O}_3{)}_2\Rightarrow \frac{1}{3}\times 5$
$\Rightarrow \frac{5}{3}=1.66mmol$
So $Pb(N{O}_3{)}_2$ is the limiting reagent.
1. 3 mmol of $Pb(N{O}_3{)}_2\Rightarrow 3$ mmol of $PbS{O}_4$
5 mmol of $Pb(N{O}_3{)}_2\Rightarrow 5mmol$
$\Rightarrow \frac{5}{1000}mol\Rightarrow 0.005mol$
so value of x is 5.