A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq). The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant and the theoretical yield.

Verified step by step guidance

1. <Calculate the moles of KCl: Use the formula \( \text{moles} = \text{molarity} \times \text{volume in liters} \) to find the moles of KCl in the 25.0 mL solution.>
2. <Calculate the moles of Pb(NO3)2: Similarly, use the formula \( \text{moles} = \text{molarity} \times \text{volume in liters} \) to find the moles of Pb(NO3)2 in the 15.0 mL solution.>
3. <Determine the limiting reactant: Use the stoichiometry of the balanced equation, 2 KCl + Pb(NO3)2 → PbCl2 + 2 KNO3, to compare the mole ratio of the reactants and identify the limiting reactant.>
4. <Calculate the theoretical yield of PbCl2: Use the moles of the limiting reactant and the stoichiometry of the reaction to calculate the moles of PbCl2 that can be formed, then convert this to grams using the molar mass of PbCl2.>
5. <Compare the theoretical yield to the actual yield: The actual yield is given as 2.45 g. Compare this to the theoretical yield to assess the efficiency of the reaction.>