The balanced chemical reaction will be as follows:
![Pb^{2+}+2I^{-}\rightarrow PbI_{2}](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%2B2I%5E%7B-%7D%5Crightarrow%20PbI_%7B2%7D)
To determine the % yield, first calculate the theoretical yield.
Molairty and volume of
is 0.218 M and 65 mL respectively. Convert it into number of moles as follows:
![n=M\times V](https://tex.z-dn.net/?f=n%3DM%5Ctimes%20V)
Here, volume should be in L thus,
![n=0.218 M\times 65\times 10^{-3}L=0.01417 mol](https://tex.z-dn.net/?f=n%3D0.218%20M%5Ctimes%2065%5Ctimes%2010%5E%7B-3%7DL%3D0.01417%20mol)
Similarly, calculate number of moles of iodide ion,
![n=0.265 M\times 80\times 10^{-3}L=0.0212 mol](https://tex.z-dn.net/?f=n%3D0.265%20M%5Ctimes%2080%5Ctimes%2010%5E%7B-3%7DL%3D0.0212%20mol)
Now, from the balanced chemical equation, 1 mole of
gives 1 mol of
, thus, 0.01417 mol will give 0.01417 mol of
.
Also, 2 mole of
will give 1 mole of
thus, 0.0212 mol will give,
![n=0.0212 mol\times \frac{1}{2}=0.0106 mol](https://tex.z-dn.net/?f=n%3D0.0212%20mol%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%3D0.0106%20mol)
Molar mass of
is 461.01 g/mol calculating mass of
obtained from
and
as follows:
From
:
![m=n\times M=0.01417 mol\times 461.01 g/mol=6.5325 g](https://tex.z-dn.net/?f=m%3Dn%5Ctimes%20M%3D0.01417%20mol%5Ctimes%20461.01%20g%2Fmol%3D6.5325%20g)
Similarly, for
:
![m=n\times M=0.0106 mol\times 461.01 g/mol=4.8867 g](https://tex.z-dn.net/?f=m%3Dn%5Ctimes%20M%3D0.0106%20mol%5Ctimes%20461.01%20g%2Fmol%3D4.8867%20g)
Here,
is limiting reactant as it produced less amount of
as compared with
.
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:
![Percentage yield=\frac{Actual yeild}{theoretical yield}\times 100](https://tex.z-dn.net/?f=Percentage%20yield%3D%5Cfrac%7BActual%20yeild%7D%7Btheoretical%20yield%7D%5Ctimes%20%20100)
Actual yield is 3.26 g thus, percentage yield will be:
=66.7%
Therefore, % yield will be 66.7%