An aqueous solution containing 5.22 g of lead(II) nitrate is added to an aqueous

Question

An aqueous solution containing 5.22 g of lead(II) nitrate is added to an aqueous solution containing 5.19 g of potassium chloride to generate solid lead(II) chloride and potassium nitrate. Write the balanced chemical equation for this reaction. Be sure to include all physical states. _______ Tip: If you need to dear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? lead(II) nitrate potassium chloride The percent yield for the reaction is 90.3%, how many grams of precipitate were recovered? Number _______ g How many grams of the excess reactant remain? Number _______ g

Explanation / Answer

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

331.2 g Pb(NO3)2 reacts with 2 x 74.55 g KCl

5.22 g Pb(NO3)2 reacts with 5.22 x 2 x 74.55 / 331.2 = 2.35 g KCl

but we have 5.19 g KCl so KCl is exess reagent and Pb(NO3)2 is limiting reagent.

limiting reagent = lead (II) nitrate

331.2 g Pb(NO3)2 forms 271.8 g PbCl2

5.22 g Pb(NO3)2 forms 5.22 x 271.8 / 331.2 = 4.28 g

% yield = 90.3

90.3 = (X / 4.28) x 100

X = 3.86 g

mass of precipitate = 3.86 g

mass of exess reagent = 5.19 - 2.35 = 2.84 g

exess reagent left = 2.84 g

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