A solid laboratory unknown contains two substances, lead(II) carbonate and potas

Question

A solid laboratory unknown contains two substances, lead(II) carbonate and potassium carbonate. A 5.000 g sample of this unknown was treated with aqueous nitric acid to dissolve it, then, after further treatment to adjust the acidity, it was treated with an excess of potassium chromate solution. The precipitate which formed, when dried carefully and weighed, has a mass of 2.697 grams. Calculate the percent, by mass, of lead in this laboratory unknown. (hint: write a balanced equation, determine the formula of the ppt using solubility chart, then grams of lead using mole ratio.)

a. 34.60%

b.50.00%

c.65.91%

d.72.93%

77.38%

Explanation / Answer

the precipitate formed is lead chromate..

Pb2+ + CrO42- => PbCrO4 (ppt)

mass of PbCrO4 = 2.697 gms

molar mass of PbCrO4 = 323.2 gms

moles of PbCrO4 = 2.697/323.2 = 0.008344 moles.

=> moles of Pb2+ = moles of PbCrO4 = 0.008344

molar mass of Pb2+ = 207.21 gms

=> mass of lead (i.e Pb2+) = 0.008344*207.21 = 1.729 gms.

there fore mass % of lead in 5 gm sample = 1.729/5*100 = 34.60%

so the answer is option A..

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