A 27.8 mL sample of a 1.20 M potassium chloride solution is mixed with 14.8 mL of a 0.870 M lead(II) nitrate solution and this precipitation reaction occurs: 2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)

Question

no of moles of KCl = molarity *volume in L= 1.20*0.0278 = 0.03336 moles KCl

no of moles of Pb(NO3)2 = molarity *volume in L = 0.87*0.0148= 0.012876 moles Pb(NO3)2

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

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

0.03336 moles KCl will react with= 0.01668 moles Pb(NO3)2 but we've just 0.012876 moles Pb(NO3)2

Pb(NO3)2 is the limiting reactant

1 mole of Pb(NO3)2 reacts with an excess of KCl to give 1 moles of PbCl2

0.012876 moles of Pb(NO3)2 react with an excess of KCl to give 0.012876 moles of PbCl2

mass of PbCl2 = no of moles *gram molar mass= 0.012876*278= 3.58 g = theoretical yield 

actual yield = 2.45g

percent yield = actual yield*100/theoretical yield= (2.45/3.58)*100= 68.4%

Answer 1

Your post reads more like a worked solution than a question. What you’ve actually done is calculate:

1. Moles of each reactant
2. Determined the limiting reagent
3. Computed the theoretical yield of PbCl₂, and
4. Found the percent yield based on an actual yield of 2.45 g.

So, what you're probably asking (though you never phrased it clearly) is likely something like:

“Can someone check my work to see if my calculations for the theoretical yield and percent yield are correct?”

or possibly

“Is my conclusion about the limiting reagent and percent yield correct?”

If I restate your problem as an actual question, it might be:

Question (rephrased):

When 27.8 mL of 1.20 M KCl reacts with 14.8 mL of 0.870 M Pb(NO₃)₂ according to

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

What is the limiting reagent, theoretical yield of PbCl₂, and percent yield if 2.45 g of PbCl₂ are actually obtained?

And yes, your reasoning and math are correct:

1. Limiting reagent: Pb(NO₃)₂
2. Theoretical yield: 3.58 g PbCl₂
3. Percent yield: 68.4 %