A 29.3-mL sample of a 1.22 M potassium chloride solution is mixed with 14.5 mL of a 0.860 M lead(II) nitrate solution and this precipitation reaction occurs:
2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)
The solid PbCl2 is collected, dried, and found to have a mass of
2.46 g. Determine the limiting reactant, the theoretical yield, and
the percent yield.
2 KCl(aq)+ Pb (NO3)2(aq)→ PbCl2(s)+ 2KNO3(aq)
1. calculate no. of moles of reactant:
KCl(aq) : Molarity = 1.22 M ;Volume = 29.3 mL , therefore no. of moles of KCl = 1.22*29.3\1000 = 0.035 mole
Pb (NO3)2 :
Molarity = 0.86 M ; Volume = 14.5 mL , therefore no. of moles of Pb (NO3)2 = 0.86*14.5\1000 = 0.012 mole
2. finding limiting reagent:
thus, limiting reactant = Pb (NO3)2 ; since in reaction it consist of lesser no. of moles i.e 0.012 mole , hich get completely used up in reaction
2. calculate theoretical yield of PbCl2
1 mole of Pb (NO3)2 = 1 mole of PbCl2
thus, 0.012 mole of Pb (NO3)2 gives 0.012 mole of PbCl2
By formula we know, no. of moles = weight \ molecular weight
Molecular weight of PbCl2 = 278 g\mol
thus 0.012 mole = weight \ 278 g\mol
therefore weight of PbCl2 = 0.012*278 =3.336 g
This is the theoretical yield of PbCl2 = 3.336 g
given: experimental yield of PbCl2 = 2.46 g
4. Calculate % yield:
by formula;
percent yield = experimental yield *100 \ theoretical yield = 2.46 *100\ 3.336 = 73.74%
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