Question

$35mL$ of $1.5MPb(N{O}_3{)}_2$ solution is mixed with $15mL$ of $0.2MKI$ solution. The maximum number of moles of $Pb{I}_2$ that can be precipitated is

• A. $5.25\times {10}^{-3}$
• B. $3\times {10}^{-3}$
• C. $1.5\times {10}^{-3}$
• D. $2.625\times {10}^{-3}$

Answer 1

The correct option is C $1.5\times {10}^{-3}$

Milli moles of $Pb(N{O}_3{)}_2=35\times 1.5=52.5$
Milli moles of $KI=15\times 0.2=3.0Pb(N{O}_3{)}_2+2KI\to Pb{I}_2+2KN{O}_3$

2 moles of $KI$ react with 1 mole of $Pb(N{O}_3{)}_2$
moles of $Pb(N{O}_3{)}_2$ react with 3 mmoles of KI = $\frac{1\times 3\times {10}^{-3}}{2}=1.5\times {10}^{-3}$
. Hence the limiting reagent is $KI$.
$\therefore \text{Moles of}Pb{I}_2\text{formed}=\frac{3}{2}\times {10}^{-3}=1.5\times {10}^{-3}$