Maximum number of mol of #PbSO_4# that can be precipitated by mixing 20.00 ml of 0.1 M #Pb(NO_3)_2# and 30.00 ml of 0.1 M #Na_2SO_4# will be?

1 Answer
Jun 19, 2017


#"0.002 moles PbSO"_4#


Start by writing the balanced chemical equation that describes this double replacement reaction

#"Pb"("NO"_ 3)_ (2(aq)) + "Na"_ 2"SO"_ (4(aq)) -> "PbSO"_ (4(s)) darr + 2"NaNO"_ (3(aq))#

Notice that the two reactants react in a #1:1# mole ratio and produce lead(II) sulfate, the precipitate, in #1:1# mole ratios.

Even without doing any calculation you should be able to say that lead(II) nitrate will act as a limiting reagent here. That happens because you're dealing with solutions of equal molarities, which implies that the solution with the bigger volume will contain more moles of solute.

To confirm this, use the molarities and volumes to find the number of moles of each reactant

#20.00 color(red)(cancel(color(black)("mL solution"))) * ("0.1 moles Pb"("NO"_ 3)_2)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0020 moles Pb"("NO"_3)_2#

#30.00color(red)(cancel(color(black)("mL solution"))) * ("0.1 moles Na"_2"SO"_4)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0030 moles Na"_2"SO"_4#

As you can see, you have fewer moles of lead(II) nitrate than of sodium sulfate, which implies that the former will act as a limiting reagent, i.e. it will be completely consumed before all the moles of sodium sulfate will get the chance to react.

You can thus use the aforementioned #1:1# mole ratios to say that the reaction consumes #0.0020# moles of lead(II) nitrate and of sodium sulfate and produces #0.0020# moles of lead(II) sulfate.

Therefore, the maximum number of moles of lead(II) sulfate that can be precipitated is equal to

#color(darkgreen)(ul(color(black)("moles PbSO"_4 = "0.002 moles")))#

The answer must be rounded to one significant figure, the number of sig figs you have for the molarities of the two solutions.