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?

The balanced chemical equation for the reaction is as follows:

From the molarity and volume of and KI, number of moles can be calculated as follows:

For :

Similarly, for KI:

From the balanced chemical reaction, 1 mol of gives 1 mole of thus, 0.01417 mol will give 0.01417 mol.

Molar mass of is 461.01 g/mol, mass can be calculated as:

Similarly, 2 mol of KI gives 1 mole of thus, 0.0212 mol will give of .

Mass can be calculated as:

The amount of obtained from KI is less than that from   thus, KI is limiting reactant and amount of obtained from KI will be theoretical yield.

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

Putting the values,

=66.80%

Therefore, % yield will be 66.80%.

the answer is barium and copper bromide