Question

Solid lead(ii) iodide was prepared by reacting 65.0 ml of a solution containing 0.218 m lead(ii) ions with 80.0 ml of a solution containing 0.265 m iodide ions. 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 reaction will be as follows:

$Pb^{2+}+2I^{-}\rightarrow PbI_{2}$

To determine the % yield, first calculate the theoretical yield.

Molairty and volume of $Pb^{2+}$ is 0.218 M and 65 mL respectively. Convert it into number of moles as follows:

$n=M\times V$

Here, volume should be in L thus,

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

Similarly, calculate number of moles of iodide ion,

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

Now, from the balanced chemical equation, 1 mole of  $Pb^{2+}$ gives 1 mol of $PbI_{2}$, thus, 0.01417 mol will give 0.01417 mol of $PbI_{2}$.

Also, 2 mole of $I^{-}$ will give 1 mole of $PbI_{2}$ thus, 0.0212 mol will give,

$n=0.0212 mol\times \frac{1}{2}=0.0106 mol$

Molar mass of $PbI_{2}$ is 461.01 g/mol calculating mass of $PbI_{2}$ obtained from  $Pb^{2+}$ and $I^{-}$ as follows:

From $Pb^{2+}$:

$m=n\times M=0.01417 mol\times 461.01 g/mol=6.5325 g$

Similarly, for $I^{-}$:

$m=n\times M=0.0106 mol\times 461.01 g/mol=4.8867 g$

Here, $I^{-}$ is limiting reactant as it produced less amount of  $PbI_{2}$  as compared with $Pb^{2+}$.

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:

$Percentage yield=\frac{Actual yeild}{theoretical yield}\times 100$

Actual yield is 3.26 g thus, percentage yield will be:

$Percentage yield=\frac{3.26}{4.8867}\times 100$=66.7%

Therefore, % yield will be 66.7%