Question

# An aqueous solution containing 5.93 g of lead(II) nitrate is added to an aqueous solution containing...

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?

Pb(NO3)2(aq.) + 2KCl (aq.) -------------> 2KNO3(aq.) + PbCl2 (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.

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