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* V`

For
`Pb(NO_(3))_(2)` :

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

Similarly, for KI:

`n=0.265 M* 80* 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* M=0.01417 mol* 461.01 g/mol=6.53 g`

Similarly, 2 mol of KI gives 1 mole of
`PbI_(2)` thus, 0.0212 mol will give
`(0.0212)/(2)=0.0106 mol` of
`PbI_(2)`.

Mass can be calculated as:

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

Putting the values,

`percentage yield=(3.26 g)/(4.88 g)* 100`=66.80%

Therefore, % yield will be 66.80%.