Final answer:
To find the number of moles of I2 formed in the reaction between KMnO4 and KI, one must use stoichiometry to analyze the balanced chemical equation and identify the limiting reagent. The molar ratios in the equation determine the amount of I2 produced based on the moles of the limiting reagent, which is KMnO4 in this case.
Step-by-step explanation:
The question asks about the reaction between potassium permanganate (KMnO4) and potassium iodide (KI) in a basic medium to form iodine (I2) and manganese dioxide (MnO2). In this reaction, KMnO4 acts as an oxidizing agent, and KI acts as a reducing agent. To calculate the number of moles of I2 formed when 250 mL of 0.1M KI reacts with 250 mL of 0.02M KMnO4, we must consider the stoichiometry of the reaction.
In the provided reaction, the stoichiometry will depend on the balanced chemical equation which is not fully given in the details, but typically, KMnO4 oxidizes I- ions to I2. However, the molar ratios between KMnO4 and I2 are crucial for the calculation. Assuming the reaction follows the typical balanced equation where 2 moles of KMnO4 react with 10 moles of KI to produce 5 moles of I2, we proceed as follows: First, determine the limiting reagent. In this case, KMnO4 limits the reaction. Then, use the stoichiometry of the balanced equation to calculate the moles of I2 based on the moles of the limiting reagent. Given these molarities and volumes, we would find that only 0.005 moles of KMnO4 are present. The correct ratio would then allow us to calculate the corresponding moles of I2.