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motikmotik
3 years ago
9

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?
Chemistry
1 answer:
professor190 [17]3 years ago
6 0

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.

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