Answer to Question #133417 in General Chemistry for Ry

Question

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%?

Answer 1

Step I : Balanced equation for first precipitation reaction

Take 1mL=1cm³

Pb(NO₃)_2(aq) + 2NaCl_(aq) $\to$ 2NaNO_3(aq) + PbCl_2(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) $\to$ PbBr_2(s) + NaNO_3(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