Question

A 29.3-mL sample of a 1.84 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs:

2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)

What is the
1. Limiting reactant
2. Theoretical yield
3. percent yield

This seems like an easy question but for some reason I keep getting hung up on these!

Answer 1

2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)

no of moles of KCl   = molarity * volume in L

                                 = 1.84*0.0293   = 0.054 moles

no of moles of Pb(NO3)2 = molarity * volume in L

                                           = 0.9*0.015    = 0.0135 moles

2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)

2 moles of KCl react with 1 moles of Pb(NO3)2

0.054 moles of KCl react with = 1*0.054/2    = 0.027 moles of Pb(NO3)2

Pb(NO3)2 is limiting reactant

1 moles of Pb(NO3)2 react with KCl to gives 1 moles of PbCl2

0.0135 moles of Pb(NO3)2 react with KCl to gives 0.0135 moles of PbCl2

mass of PbCl2 = no of moles * gram molar mass

                            = 0.0135*278   = 3.753g

Theoritical yield of PbCl2 = 3.753g

percentage yiled = actual Yield*100/theoritical yield

But in this proble doest not given the actual yield