Question

An aqueous solution containing 5.93 g of lead(II) nitrate is added to an aqueous solution containing 5.88 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. How many grams of the excess reactant remain?

Answer 1

Pb(NO₃)_2(aq.) + 2KCl _(aq.) -------------> 2KNO_3(aq.) + PbCl_{2 (s)}

Mass of Pb(NO₃)₂ = 5.93 g

Molecular weight of Pb(NO₃)₂ = 331.2 g/mol

no. of moles of Pb(NO₃)₂ = 5.93/331.2 g/mol = 0.0179 mol

Mass of KCl = 5.88 g

Molecular weight of KCl = 74.55 g/mol

no. of moles of KCl = 5.88/74.55 g/mol = 0.0788 mol

One mole of Pb(NO₃)₂ reacts with two moles of KCl to give two moles of KNO₃ and one mole of PbCl₂

0.0179 mol  of Pb(NO₃)₂ reacts with 0.0358 moles of KCl to give 0.0358 moles of KNO₃ and 0.0179 mole of PbCl₂

Limiting Reagent:: Pb(NO₃)₂

Excess Reagent: KCl

Excess Reagent (KCl) is 0.0788 - 0.0358 = 0.0430 moles

0.0430 moles of KCl corresponds to 0.0430 mol x 74.55g.mol = 3.206g

3.206 grams of the excess reactant remain in solution.