Question

When 11.25 moles of lead nitrate are made to react with 2.5 moles of chromic sulphate, precipitation of lead sulphate takes place. Calculate the moles of $Cr(N{O}_3{)}_3$ formed in the final solution. Assume that lead sulphate is completely insoluble.
(Molar mass of lead - 207 g/mol)

• A. 7
• B. 3
• C. 5
• D. 2

Answer 1

The correct option is C 5

$3Pb(N{O}_3{)}_2+C{r}_2(S{O}_4{)}_3\to 3PbS{O}_4+2Cr(N{O}_3{)}_3$
Initial moles of $Pb(N{O}_3{)}_2=11.25$ moles
Initial moles of $C{r}_2(S{O}_4{)}_3=2.5$ moles
Calculating limiting reagent:
3 moles of $Pb(N{O}_3{)}_2$ react with 1 mole of $C{r}_2(S{O}_4{)}_3$
11.25 moles of $Pb(N{O}_3{)}_2$will react with $\frac{1}{3}\times 11.25=3.75$ moles of $C{r}_2(S{O}_4{)}_3$
Hence, the limiting reagent is chromic sulphate.
1 mole of $C{r}_2(S{O}_4{)}_3$ produces 2 moles of $Cr(N{O}_3{)}_3$
2.5 moles of $C{r}_2(S{O}_4{)}_3$ will produce $2\times 2.5=5$ moles of $Cr(N{O}_3{)}_3$
So, moles of $Cr(N{O}_3{)}_3$ produced $=2\times 2.5=5$ moles