To determine the amount of lead II iodide that could be formed, we need to first calculate the limiting reagent in the reaction. This is the reactant that is completely consumed and limits the amount of product that can be formed.
The balanced chemical equation for the reaction is:
Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3
From the equation, we can see that 2 moles of KI are required to react with 1 mole of Pb(NO3)2 to form 1 mole of PbI2.
The molar mass of KI is 166.00 g/mol, so 25.00 grams of KI is equivalent to:
25.00 g / 166.00 g/mol = 0.1506 mol KI
Based on the balanced chemical equation, we know that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to form 1 mole of PbI2. Therefore, the number of moles of Pb(NO3)2 required to react with 0.1506 mol KI is:
0.1506 mol KI × (1 mol Pb(NO3)2 / 2 mol KI) = 0.0753 mol Pb(NO3)2
The molar mass of PbI2 is 461.01 g/mol, so the mass of PbI2 that could be formed from 0.0753 mol Pb(NO3)2 is:
0.0753 mol Pb(NO3)2 × (1 mol PbI2 / 1 mol Pb(NO3)2) × 461.01 g/mol = 34.76 g PbI2
Therefore, if we start with 25.00 grams of potassium iodide, we could form a maximum of 34.76 grams of lead II iodide.