Question 28 of 34 Sapling Learning Map An aqueous solution containing 6.03 g of

Question

Question 28 of 34 Sapling Learning Map An aqueous solution containing 6.03 g of lead(l) nitrate is added to an aqueous solution containing 6.28 g of Enter the balanced chemical equation for this reaction. Be sure to include all physical states potassium chloride. Tipe yo nd t cdear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? O lead() nitrate O potassium chloride The percent yield for the reaction is 84.0%, how many grams of precipitate were recovered? Number How many grams of the excess reactant remain? Number Exit Previous > Give Up & View Solution e Check Answer 0 Next

Explanation / Answer

Part a

The balanced reaction

2KCl(aq) + Pb(NO3)2(aq) ===> PbCl2(s) + 2KNO3(aq)

MW (2*74.55) (331.2)    (278.1) (2*101.1)

Part b

From the stoichiometry of the reaction

(2*74.55) g KCl reacts with = 331.2 g Pb(NO3)2

6.28 g KCl reacts with = 331.2*6.28/(2*74.55)

= 13.95 g Pb(NO3)2

But we have only 6.03 g Pb(NO3)2

Limiting reactant = Lead (II) Nitrate

Part c

Precipitate is PbCl2

Theoretical yield of PbCl2

= 278.1 g PbCl2 x 6.03 g Pb(NO3)2 / 331.2 g Pb(NO3)2

= 5.063 g PbCl2

Actual yield = % yield x theoretical yield

= 0.84 x 5.063

= 4.25 g

Precipitate recovered = 4.25 g

Part d

Excess reactant (KCl) remain = initially present - reacted

= 6.28 - (2*74.55 g KCl *6.03 g Pb(NO3)2 /331.2 g Pb(NO3)2)

= 6.28 - 2.71

= 3.57 g

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