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* V`

Here, volume should be in L thus,

`n=0.218 M* 65* 10^(-3)L=0.01417 mol`

Similarly, calculate number of moles of iodide ion,

`n=0.265 M* 80* 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* (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* M=0.01417 mol* 461.01 g/mol=6.5325 g`

Similarly, for
`I^(-)`:

`m=n* M=0.0106 mol* 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=(Actual yeild)/(theoretical yield)*  100`

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

`Percentage yield=(3.26)/(4.8867)*  100`=66.7%

Therefore, % yield will be 66.7%