Question #05ae8

The idea here is that because the precipitation was complete, all the lead that was present in the initial solution will now be part of the precipitate, which in this case is lead(II) sulfide, #"PbS"#.

You don't actually need a balanced chemical equation to solve this problem, all you have to do is use the concept of mass conservation.

Calculate the percent composition of lead in lead(II) sulfide by using the compound's molar mass

#"% Pb in PbS" = (207 color(red)(cancel(color(black)("g mol"^(-1)))))/((207 + 32)color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = 86.61%#

Do the same for lead(II) nitrate, #"Pb"("NO"_3)_2#

#"% Pb in Pb"("NO"_3)_2 = (207 color(red)(cancel(color(black)("g mol"^(-1)))))/((207 + 2 * 14 + 6 * 16)color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = 62.54%#

Now, use the percent composition of the lead(II) sulfide to find how much lead in the form of lead(II) cations, #"Pb"^(2+)#, you have in the precipitate

#0.200 color(red)(cancel(color(black)("g PbS"))) * "86.61 g Pb"/(100color(red)(cancel(color(black)("g PbS")))) = "0.17322 g Pb"#

This is exactly how much lead you had in the initial sample of lead(II) nitrate, so use this value to find the actual mass of lead(II) nitrate in the initial sample

#0.17322 color(red)(cancel(color(black)("g Pb"))) * ("100 g Pb"("NO"_3)_2)/(62.54color(red)(cancel(color(black)("g Pb")))) = "0.277 g Pb"("NO"_3)_2#

Since the initial sample had mass of #"0.331 g"#, it follows that its purity was approximately equal to

#("0.277 g Pb"("NO"_3)_2)/"0.331 g sample" xx 100 = 83.7% ~~ color(green)(|bar(ul(color(white)(a/a)color(black)(84%)color(white)(a/a)|)))#

SIDE NOTE Notice how much extraneous information is provided in the problem!

You don't need to know the volume of the solution. Likewise, you don't need to know that the sample is contaminated with sodium nitrate.

All you need to know here is the mass of the precipitate and the mass of the initial sample, provided of course that the precipitation is complete.

ALTERNATIVELY

You can double-check your result by using moles. The balanced chemical equation for this reaction looks like this

#"Pb"("NO"_ 3)_ (2(aq)) + "H"_ 2"S"_ ((g)) -> "PbS"_ ((s)) darr + 2"HNO"_ (3(aq))#

Notice that one mole of lead(II) nitrate produces one mole of lead(II) sulfide.

Use the molar mass of lead(II) sulfide to find the number of moles produced

#0.200 color(red)(cancel(color(black)("g"))) * ("1 mole PbS")/((207 + 32)color(red)(cancel(color(black)("g")))) = "0.0008368 moles PbS"#

This means that the reaction consumed #0.0008368# moles of lead(II) nitrate, which is equivalent to

#0.0008368color(red)(cancel(color(black)("moles Pb"("NO"_3)_2))) * ((207 + 2 * 14 + 6 * 16)color(white)(a)"g")/(1color(red)(cancel(color(black)("mole Pb"("NO"_3)_2)))) = "0.277 g"#

Once again, you have

#("0.277 g Pb"("NO"_3)_2)/"0.331 g sample" xx 100 = 83.7% ~~ color(green)(|bar(ul(color(white)(a/a)color(black)(84%)color(white)(a/a)|)))#