Answer :
Answer: 2.78 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
[tex]\text{no of moles}of FeCl_3={0.628M}\times {0.0163 L}=0.010moles[/tex]
[tex]\text{no of moles}of Pb(NO_3)_2={0.642M}\times 0.0165L}=0.010moles[/tex]
[tex]2FeCl_3(aq)+3Pb(NO_3)_2(aq0\rightarrow 3PbCl_2(s)+2Fe(NO_3)_3(aq)[/tex]
According to stoichiometry:
3 moles of [tex]Pb(NO_3)_2[/tex] reacts with 2 moles of [tex]FeCl_3[/tex]
Thus 0.010 moles of [tex]Pb(NO_3)_2[/tex] will give=[tex]\frac{2}{3}\times 0.010=0.006moles[/tex] of [tex]FeCl_3[/tex]
[tex]Pb(NO_3)_2[/tex] is a limiting reagent as it limits the formation of products and[tex]FeCl_3[/tex] is an excess reagent.
As 3 moles of [tex]Pb(NO_3)_2[/tex] give = 3 moles of [tex]PbCl_2[/tex]
Thus 0.010 moles of [tex]Pb(NO_3)_2[/tex] will give=[tex]\frac{3}{3}\times 0.010=0.010moles[/tex] of [tex]PbCl_2[/tex]
mass of [tex]PbCl_2=moles\times {\text {molar mass}}=0.010\times 278 = 2.78grams[/tex]
Thus the theoretical yield (in g of precipitate) when 16.3 mL of a 0.628 M solution of iron(III) chloride is combined with 16.5 mL of a 0.642 M solution of lead(II) nitrate is 2.78 grams