Answer: Approximately 57.23%
Explanation: To calculate the percentage yield of lead iodide, we need to compare the experimental yield (the amount of product actually obtained) to the theoretical yield (the amount of product that would be obtained if the reaction went to completion). The balanced chemical equation for the reaction that produces lead iodide is: Pb(NO3)2 + 2KI → PbI2 + 2KNO3 From this equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2. Therefore, the theoretical yield of PbI2 can be calculated as follows: Theoretical yield of PbI2 = 0.005 moles Pb(NO3)2 x (1 mole PbI2 / 1 mole Pb(NO3)2) x (331.2 g PbI2 / 1 mole PbI2) = 1.66 g PbI2 where 331.2 g/mol is the molar mass of PbI2. Now we can calculate the percentage yield of PbI2 as follows: Percentage yield = (experimental yield / theoretical yield) x 100% Percentage yield = (0.95 g PbI2 / 1.66 g PbI2) x 100% = 57.23% Therefore, the percentage yield of lead iodide is approximately 57.23%.