Question

Given the reaction:-

$Pb{\left(N{O}_3\right)}_2\left(aq\right)+2KI\to Pb{I}_2\left(s\right)+2KN{O}_3\left(aq\right)$ What is the mass of $Pb{I}_2$ that will precipitate if 10.2 g of $Pb{\left(N{O}_3\right)}_2$ is mixed with 5.73 g of $KI$ in a sufficient quantity of $H_{2}O$ ?

• A. 2.06 g
• B. 4.13 g
• C. 7.96 g
• D. 15.9 g

Answer 1

The correct option is C 7.96 g

$Pb(N{O}_3{)}_2+2KI\to Pb{I}_2+2KN{O}_3$

$331.2g$ $2\times 166$ $461$

$10.2g$ $5.73g$ $?$

$KI$ is limiting reagent, So it will decide the product.

$\Rightarrow 5.73g$ of $KI$ gives $=\frac{461}{332}\times 5.73$

$=7.96g$

So, the correct option is $C$