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?

The balanced chemical reaction will be as follows:

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

Molairty and volume of is 0.218 M and 65 mL respectively. Convert it into number of moles as follows:

Here, volume should be in L thus,

Similarly, calculate number of moles of iodide ion,

Now, from the balanced chemical equation, 1 mole of   gives 1 mol of , thus, 0.01417 mol will give 0.01417 mol of .

Also, 2 mole of will give 1 mole of thus, 0.0212 mol will give,

Molar mass of is 461.01 g/mol calculating mass of obtained from   and as follows:

From :

Similarly, for :

Here, is limiting reactant as it produced less amount of    as compared with .

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:

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

=66.7%

Therefore, % yield will be 66.7%

the answer is c. nacl

you gotta add the following molecules so we can give u an answer

explanation: