Final answer:
The percent yield of lead(II) iodide is calculated by dividing the actual yield of 26.7 grams by the theoretical yield of 86.44 grams and then multiplying by 100%, resulting in a percent yield of approximately 30.89%.
Step-by-step explanation:
To calculate the percent yield of lead(II) iodide, first determine the theoretical yield based on stoichiometry. For the reaction 2KI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2KNO₃(aq), a 750 mL solution of 0.250 M KI contains 0.1875 moles of KI (since 0.750 L × 0.250 mol/L = 0.1875 mol). Based on the reaction stoichiometry, 1 mole of Pb(NO₃)₂ reacts with 2 moles of KI, so the theoretical yield of PbI₂ corresponds to the moles of KI divided by 2.
The molar mass of PbI₂ is 461.01 g/mol. Thus, the theoretical yield in grams would be the moles of KI (0.1875 mol) × (461.01 g/mol), which is approximately 86.44 g of PbI₂. With the actual yield being 26.7 g, the percent yield is calculated as:
Percent Yield = (Actual Yield / Theoretical Yield) × 100%
Percent Yield = (26.7 g / 86.44 g) × 100%
Percent Yield ≈ 30.89%