An aqueous solution containing 7.01 g of lead(II) nitrate is added to an aqueous

Question

An aqueous solution containing 7.01 g of lead(II) nitrate is added to an aqueous solution containing 5.46 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? lead(II) nitrate O potassium chloride The percent yield for the reaction is 81.3%, how many grams of precipitate were recovered? Number How many grams of the excess reactant remain? Number Previous Give Up &View; Solution Check Answer Next E

Explanation / Answer

Q1.

m = 7.01 g of Pb(NO3)2

m = 5.46 g of KCl

the reaction:

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

now

MW of Pb(NO3)2 = 331.2098

MW of KCl = 74.5513

mol of Pb+2 = mass/MW = 7.01/331.2098 = 0.02116

mol of Cl- = mass/MW = 5.46/74.5513 = 0.07323

ratio is 1:2 so

Pb+2 is limiting, since it can only form --> 0.02116 mol of Pb+2 = 0.02116 mol of PbCl2

choose Lead(II) nitrate

c)

if only 81.3% is recovered

mass of PbCl2 possible = mol*MW = 0.02116 *278.1 = 5.884596 g

at 81.3% yield --> 81.3/100*5.884596 = 4.78417 g

d)

excess reactant -->0.02116 *0.813 = 0.01720 mol of Pb+2 reacts

mol of KCl reacted = 2*0.01720 = 0.0344

mass of Kcl reacted = mol*MW = 0.0344*74.551 = 2.564 g

mass of KCl elft = 5.46-2.564 = 2.896 g of KCL

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