Question

Consider the following balanced equation for the precipitation reaction that occurs in aqueous solution.

Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)

Answer the following questions if 0.50 L of 0.40 M Pb(NO₃)₂ is mixed with 0.50 L of 0.40 M KI:

Part A - How many moles of Pb(NO3)2 are present in solution?

Part B - How many moles of KI are present in this solution?

Part D - How many moles of PbI2 can form based upon complete reaction of Pb(NO3)2 and KI ?

Part E - What mass of the precipitate (the solid product) should be expected to form in this reaction? (What is the theoretical yield of PbI2?)

Answer 1

Answer – We are given, reaction –

Pb(NO3)2(aq)+2KI(aq) ------> PbI₂(s)+2KNO3(aq)

0.50 L of 0.40 M Pb(NO₃)₂ is mixed with 0.50 L of 0.40 M KI

Part A) Moles of Pb(NO₃)₂ = 0.40 M * 0.50 L

                                             = 0.20 moles

Part B) Moles of KI = 0.40 M * 0.50 L

                                = 0.20 moles

Part C) moles of PbI₂(s)

From the above reaction

1 mole of Pb(NO₃)₂ = 1 moles of PbI₂(s)

So, 0.20 moles of Pb(NO₃)₂ = ?

= 0.20 moles of PbI₂(s)

From KI –

2 mole of KI = 1 moles of PbI₂(s)

So, 0.20 moles of KI = ?

= 0.10 moles of PbI₂(s)

Part E - theoretical yield of PbI2

From the above part we found that KI is limiting reactant and the

moles of PbI₂(s) = 0.10 moles

so, mass of PbI₂(s) = 0.10 moles * 461 g/mol

                              = 46.1 g