Answer to Question #133417 in General Chemistry for Ry

Question #133417
A 185 mL sample of 1.20 M Pb(NO₃)₂ is mixed with 82.5 mL of 1.50 M NaCl, and the PbCl₂ precipitate is filtered from the solution. Then 200 mL of 3.00 M NaBr is added to the remaining solution, and the PbBr₂ precipitate is also collected and dried. What is the mass of the PbBr₂ precipitate, assuming the yield in each precipitation step is 100%?
1
Expert's answer
2020-09-17T07:17:26-0400

Step I : Balanced equation for first precipitation reaction

Take 1mL=1cm3


Pb(NO3)2(aq) + 2NaCl(aq) "\\to" 2NaNO3(aq) + PbCl2(s)


Moles of Pb(NO3)2 that reacted

1.20mol were in 1000cm3

? moles would be in 185cm3

"=\\dfrac{1.2molx185cm^3}{1000cm^3} = 0.222moles"


Moles of NaCl present were


1.50 moles were in 1000cm3

? moles would be in 85.5cm3


"= \\dfrac{1.50mol x85.5cm^3}{1000cm^3} = 0.12825 moles"


Moles of Pb(NO3)2that reacted with NaCl are half of the moles of NaCl from the mole ratios;

Thus, they are"=\\dfrac{1}{2}x0.12825 = 0.065125 moles"


Therefore moles of Pb(NO3)2 were in excess

Moles of Pb(NO3)2 remaining would be?

=0.222moles - 0.06425moles

=0.157875 moles


These are the moles reacted with NaBr in the second precipitation reaction


Step II: Balanced equation for the second precipitation


Pb(NO3)2(aq) + 2NaBr(aq) "\\to" PbBr2(s) + NaNO3(aq)


From the look of things Pb(NO3)2 is the limiting factor in this reaction.


Therefore moles of PbBr2 produced is equal to the moles of Pb(NO3)2 that reacted.

i.e. mole ratio is 1:1


Thus moles of PbBr2 produced"=\\dfrac{1}{1}x0.157875 = 0.157875moles"



Mass of PbBr2 precipitate is = Moles x RMM ( use RAM of Pb=207.2, Br=79.9)


RMM of PbBr2 = 207.2 + (2 x 79.9) = 367.0g/mol


Mass = 0.157875 moles x 367.0g/mol


= 57.940g



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