Question

1. You are given 4.65g Lead (II) Nitrate and you mix it with 35ml of 0.60M Potassium lodide. (10 marks) A) Write a Balanced Chemical Equation showing reactants and products. (1 mark) B) Determine the number of moles of each reactant (2 marks) C) Determine the Limiting and Excess Reactants (1 marks) D) Determine the Mass of Excess (2 marks) E) Determine the mass of the mass of Lead (II) iodide (2 marks) F) If you form 4.75 g of Lead (II) iodide determine the Percent Yield. (2 marks)

Answer 1

A) Pb(NO3) 2 (aq) + 2 KI (aq) $\rightarrow$ PbI2 (s) + 2 KNO 3 (aq)  

B)

Mass of Pb(NO3) 2 = 4.65 g

Molar Mass of Pb(NO3) 2 = 331.20 g/mol

Moles of Pb(NO3) 2 = Mass / Molar mass = 4.65 g / ( 331.20 g/mol ) = 0.01404 mol

We have relation, Molarity = Moles of solute / Volume of solution in L

Moles of KI = Molarity $\times$ Volume of solution in L

Moles of KI = 0.60 mol / L $\times$ 0.035 L = 0.021 mol

C )

Consider reaction, Pb(NO3) 2 (aq) + 2 KI (aq) $\rightarrow$ PbI2 (s) + 2 KNO 3 (aq)  

According to reaction, 1 mol Pb(NO3) 2 reacts with 2 mol KI. Hence, moles of KI required to react with 0.01404 mol  Pb(NO3) 2 would be

0.01404 mol  Pb(NO3) 2 $\times$ ( 2 mol KI / 1 mol  Pb(NO3) 2 ) = 0.02808 mol KI

Provided moles of KI ( 0.021 ) are less than required moles of KI ( 0.02808 ) . Hence, KI is limiting reactant.

Pb(NO3) 2 is excess reactant.

D)

According to reaction, 1 mol Pb(NO3) 2 reacts with 2 mol KI.  

Hence, moles of Pb(NO3) 2 consumed in the reaction would be

0.021 mol KI $\times$ ( 1 mol  Pb(NO3) 2 / 2 mol KI ) = 0.0105 mol  Pb(NO3) 2

Moles of unreacted  Pb(NO3) 2 = 0.01404 - 0.0105 = 0.00354 mol

Mass of unreacted  Pb(NO3) 2 = 0.00354 mol $\times$ ( 331.20 g/mol ) = 1.17 g

ANSWER : Excess mass = 1.17 g

E)

Accoridng to reaction, 2 mol KI produces 1 mol PbI2 .

Hence, moles of PbI2 produced in the reaction are

0.021 mol KI $\times$ ( 1 mol PbI2 / 2 mol KI ) = 0.0105 mol PbI2

Mass of  PbI2 produced in the reaction = 0.0105 mol $\times$ 461.01 g / mol = 4.84 g

F) % yield = ( Actual yield / Theoretical yield ) $\times$ 100

% yield = ( 4.75 g / 4.84 g ) $\times$ 100

= 98.1 %