Question

An aqueous solution containing 5.01 g of lead(ii) nitrate is added to an aqueous solution containing 5.63 g of potassium chloride. enter the balanced chemical equation for this reaction. be sure to include all physical states. balanced chemical equation: Pb(NO3)2(aq) + 2KCl(aq) ⟶ PbCl₂(s) + 2KNO3(aq) what is the limiting reactant? potassium chloride lead(ii) nitrate the percent yield for the reaction is 80.5% . how many grams of the precipitate are formed? precipitate formed: g taking into account the percent yield, how many grams of the excess reactant (the reactant that is not limiting) remain? excess reactant remaining: A. Excess reactant: Potassium chloride, Remaining amount: 0.52 g B. Excess reactant: Lead(II) nitrate, Remaining amount: 0.59 g C. Excess reactant: Potassium chloride, Remaining amount: 1.11 g D. Excess reactant: Lead(II) nitrate, Remaining amount: 1.25 g

Answer 1

To find the limiting reactant between lead(II) nitrate and potassium chloride, we convert the given masses to moles and compare them to the balanced reaction's stoichiometry. The reactant with the lesser mole ratio is limiting, and it determines the theoretical and actual (considering percent yield) amount of precipitate formed. Then, the remaining excess reactant mass is calculated.

Step-by-step explanation:

The chemical equation for the reaction between aqueous lead(II) nitrate and aqueous potassium chloride is:

Pb(NO3)2(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO3(aq)

To determine the limiting reactant, we need to compare the mole ratio between lead(II) nitrate and potassium chloride with the stoichiometry of the balanced chemical equation. This calculation involves converting the masses given to moles using their respective molar masses. Once we've identified the limiting reactant, we use it to determine the theoretical yield of the precipitate (lead(II) chloride) formed, apply the percent yield given (80.5%), and calculate the actual yield. We can then calculate the remaining mass of the excess reactant.