Question

Lead(ii) iodide was prepared by reacting 65.0 ml of 0.218 m pb(no3)2 with 80.0 ml of 0.265 m ki. If the actual yield of the reaction was 3.26 g, which choice is closest to the %yield of the reaction?

Answer 1

The balanced chemical equation for the reaction is as follows:

$Pb(NO_{3})_{2}+2KI\rightarrow PbI_{2}+2KNO_{3}$

From the molarity and volume of $Pb(NO_{3})_{2}$ and KI, number of moles can be calculated as follows:

$n=M\times V$

For $Pb(NO_{3})_{2}$ :

$n=0.218 M\times 65\times 10^{-3}L=0.01417 mol$

Similarly, for KI:

$n=0.265 M\times 80\times 10^{-3}L=0.0212 mol$

From the balanced chemical reaction, 1 mol of $Pb(NO_{3})_{2}$ gives 1 mole of $PbI_{2}$ thus, 0.01417 mol will give 0.01417 mol.

Molar mass of $PbI_{2}$ is 461.01 g/mol, mass can be calculated as:

$m=n\times M=0.01417 mol\times 461.01 g/mol=6.53 g$

Similarly, 2 mol of KI gives 1 mole of $PbI_{2}$ thus, 0.0212 mol will give $\frac{0.0212}{2}=0.0106 mol$ of $PbI_{2}$.

Mass can be calculated as:

$m=n\times M=0.0106 mol\times 461.01 g/mol=4.88 g$

The amount of $PbI_{2}$ obtained from KI is less than that from  $Pb(NO_{3})_{2}$ thus, KI is limiting reactant and amount of $PbI_{2}$ obtained from KI will be theoretical yield.

Actual yield is 3.26 g, % yield can be calculated as follows:

$percentage yield=\frac{Actual yield}{theoretical yield}\times 100$

Putting the values,

$percentage yield=\frac{3.26 g}{4.88 g}\times 100$=66.80%

Therefore, % yield will be 66.80%.