Final answer:
A 0.660 L of 0.359 M KI solution provides enough reactant to form 0.11847 moles of PbI₂ when added to an excess Pb(NO₃)₂ solution, following a simple stoichiometric calculation based on the balanced equation provided.
Step-by-step explanation:
The question asks if PbI₂ can be formed by adding 0.660 L of a 0.359 M solution of KI to a solution with an excess of Pb(NO₃)₂. The reaction follows the equation:
2KI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2KNO₃(aq)
According to this balanced chemical equation, two moles of KI react with one mole of Pb(NO₃)₂ to form lead iodide (PbI₂), which is a precipitate. The solubility guidelines confirm that PbI₂ is the only insoluble compound formed, hence it precipitates out. To find out how much PbI₂ is formed, we first calculate the number of moles of KI:
0.660 L × 0.359 M = 0.23694 moles of KI
Since the stoichiometry is 2 moles KI to 1 mole PbI₂, therefore:
0.23694 moles KI × (1 mole PbI₂ / 2 moles KI) = 0.11847 moles PbI₂
Considering the molar mass of PbI₂, we can calculate the mass of the precipitate:
0.11847 moles × Molar mass of PbI₂ = Mass of PbI₂ precipitated