Question

Consider the balanced equation of K I KI reacting with P b ( N O 3 ) 2 Pb(NOX3)X2 to form a precipitate. 2 K I ( a q ) + P b ( N O 3 ) 2 ( a q ) ⟶ P b I 2 ( s ) + 2 K N O 3 ( a q ) 2KI(aq)+Pb(NOX3)X2(aq)⟶PbIX2(s)+2KNOX3(aq) What mass of P b I 2 PbIX2 can be formed by adding 0.528 L of a 0.417 M solution of K I KI to a solution of excess P b ( N O 3 ) 2 Pb(NOX3)X2?

Answer 1

Answer: 50.7 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $KI$

$\text{Number of moles}=molarity\times {\text {Volume in L]}=0.417M\times 0.528L=0.220moles$

The balanced chemical equation is:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

$KI$ is the limiting reagent as it limits the formation of product and $Pb(NO_3)_2$ is in excess.

According to stoichiometry :

2 moles of $KI$ give =  1 mole of $PbI_2$

Thus 0.220 moles of $KI$ give=$\frac{1}{2}\times 0.220=0.110moles$  of $PbI_2$

Mass of $PbI_2=moles\times {\text {Molar mass}}=0.110moles\times 461.01g/mol=50.7g$

Thus 50.7 g of $PbI_2$ will be formed.