Part A Lead ions can be precipitated from solution with KCI according to the fol

Question

Part A Lead ions can be precipitated from solution with KCI according to the following reaction: Pb2 + (aq) + 2 KCI(aq) PbCle(s) + 2K+ (aq) When 28.7 g KCl is added to a solution containing 25.9 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.1 g Determine the limiting reactant o KCI Submit My Answers Give Up Part B Determine the theoretical yield of PbCl2 Submit My Answers Give Up Part C Determine the percent yield for the reaction. Submit My Answers Give Up

Explanation / Answer

A)

Molar mass of Pb = 207.2 g/mol

mass of Pb = 25.9 g

we have below equation to be used:

number of mol of Pb,

n = mass of Pb/molar mass of Pb

=(25.9 g)/(207.2 g/mol)

= 0.125 mol

Molar mass of KCl = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass of KCl = 28.7 g

we have below equation to be used:

number of mol of KCl,

n = mass of KCl/molar mass of KCl

=(28.7 g)/(74.55 g/mol)

= 0.385 mol

we have the Balanced chemical equation as:

Pb2+ + 2 KCl ---> PbCl2 + 2 K+

1 mol of Pb reacts with 2 mol of KCl

for 0.125 mol of Pb, 0.25 mol of KCl is required

But we have 0.385 mol of KCl

so, Pb2+ is limiting reagent

Answer: Pb2+

B)

we will use Pb2+ in further calculation

From balanced chemical reaction, we see that

when 1 mol of Pb reacts, 1 mol of PbCl2 is formed

mol of PbCl2 formed = (1/1)* moles of Pb

= (1/1)*0.125

= 0.125 mol

  

Molar mass of PbCl2 = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

we have below equation to be used:

mass of PbCl2 = number of mol * molar mass

= 0.125*2.781*10^2

= 34.76 g

Answer: 34.8 g

C)

% yield = actual mass*100/theoretical mass

= 29.1*100/34.76

= 83.7 %

Answer: 83.7 %

Feel free to comment below if you have any doubts or if this answer do not work

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