The balanced chemical equation for the precipitation reaction of lead nitrate Pb NO3 2 and potassium iodide KI is:Pb NO3 2 aq + 2 KI aq PbI2 s + 2 KNO3 aq The product formed is lead iodide PbI2 , which is a yellow precipitate.To determine how much of this product would be produced from 25 grams of each reactant, we need to perform stoichiometric calculations.First, we need to find the molar mass of each reactant and product:Pb NO3 2: 207.2 Pb + 2 * 14.01 N + 3 * 16.00 O = 331.21 g/molKI: 39.10 K + 126.90 I = 166.00 g/molPbI2: 207.2 Pb + 2 * 126.90 I = 460.99 g/molNext, we need to convert the mass of each reactant to moles:moles of Pb NO3 2 = 25 g / 331.21 g/mol = 0.0755 molmoles of KI = 25 g / 166.00 g/mol = 0.1506 molNow, we need to determine the limiting reactant:Pb NO3 2 to KI mole ratio = 1:20.0755 mol Pb NO3 2 * 2 mol KI / 1 mol Pb NO3 2 = 0.151 mol KI requiredSince we have 0.1506 mol of KI, KI is the limiting reactant.Finally, we can calculate the moles of PbI2 produced and convert it to grams:moles of PbI2 produced = 0.1506 mol KI * 1 mol PbI2 / 2 mol KI = 0.0753 molmass of PbI2 produced = 0.0753 mol * 460.99 g/mol = 34.71 gSo, 34.71 grams of lead iodide PbI2 would be produced from 25 grams of each reactant.