Question

On mixing $45.0$ mL of $0.25M$ lead nitrate solution with $25.0$ mL of $0.10M$ chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Assume that lead sulphate is completely insoluble.

• A. $1$
• B. $7.5$
• C. $3.5$
• D. $4$

Answer 1

The correct option is B $7.5$

The balanced reaction is as follows:

$3Pb(N{O}_3{)}_2+C{r}_2(S{O}_4{)}_3\to 3PbS{O}_4\downarrow +2Cr(N{O}_3{)}_3$
The number of mmol of $Pb(N{O}_3{)}_2$ $=45.0\times 0.25=11.25$
The number of mmol of $C{r}_2(S{O}_4{)}_3$ $=25.0\times 0.10=2.5$
$C{r}_2(S{O}_4{)}_3$ is the limiting reagent.
$7.5$ mmol of $Pb(N{O}_3{)}_2$ will react with $2.5$ mmol of $C{r}_2(S{O}_4{)}_3$ to form $7.5$ mmol of $PbS{O}_4$.
$11.25-7.5=3.75$ mmol of $Pb(N{O}_3{)}_2$ will remain unreacted.

Its concentration will be $\frac{3.75}{45.0+25.0}=0.0536M$