Respuesta :
The balanced chemical reaction will be as follows:
[tex]Pb^{2+}+2I^{-}\rightarrow PbI_{2}[/tex]
To determine the % yield, first calculate the theoretical yield.
Molairty and volume of [tex]Pb^{2+}[/tex] is 0.218 M and 65 mL respectively. Convert it into number of moles as follows:
[tex]n=M\times V[/tex]
Here, volume should be in L thus,
[tex]n=0.218 M\times 65\times 10^{-3}L=0.01417 mol[/tex]
Similarly, calculate number of moles of iodide ion,
[tex]n=0.265 M\times 80\times 10^{-3}L=0.0212 mol[/tex]
Now, from the balanced chemical equation, 1 mole of [tex]Pb^{2+}[/tex] gives 1 mol of [tex]PbI_{2}[/tex], thus, 0.01417 mol will give 0.01417 mol of [tex]PbI_{2}[/tex].
Also, 2 mole of [tex]I^{-}[/tex] will give 1 mole of [tex]PbI_{2}[/tex] thus, 0.0212 mol will give,
[tex]n=0.0212 mol\times \frac{1}{2}=0.0106 mol[/tex]
Molar mass of [tex]PbI_{2}[/tex] is 461.01 g/mol calculating mass of [tex]PbI_{2}[/tex] obtained from [tex]Pb^{2+}[/tex] and [tex]I^{-}[/tex] as follows:
From [tex]Pb^{2+}[/tex]:
[tex]m=n\times M=0.01417 mol\times 461.01 g/mol=6.5325 g[/tex]
Similarly, for [tex]I^{-}[/tex]:
[tex]m=n\times M=0.0106 mol\times 461.01 g/mol=4.8867 g[/tex]
Here, [tex]I^{-}[/tex] is limiting reactant as it produced less amount of [tex]PbI_{2}[/tex] as compared with [tex]Pb^{2+}[/tex].
Theoretical yield is amount of product obtained from limiting reactant thus, theoretical yield will be 4.8867 g.
Percentage yield can be calculated as follows:
[tex]Percentage yield=\frac{Actual yeild}{theoretical yield}\times 100[/tex]
Actual yield is 3.26 g thus, percentage yield will be:
[tex]Percentage yield=\frac{3.26}{4.8867}\times 100[/tex]=66.7%
Therefore, % yield will be 66.7%
