Question #6ca13

1 Answer
May 3, 2017


#"3.22 g"#


For starters, you need a balanced chemical equation to go by

#2"FeCl"_ (3(aq)) + 3"Pb"("NO"_ 3)_ (2(aq)) -> 3"PbCl"_ (2(s)) darr + 2"Fe"("NO"_ 3)_ (3(aq))#

Use the molarities and the volumes of the two solutions to figure out how many moles of each reactant are available.

#16.9 color(red)(cancel(color(black)("mL"))) * "0.582 moles FeCl"_2/(10^3color(red)(cancel(color(black)("mL")))) = "0.009836 moles FeCl"_3#

#15.5 color(red)(cancel(color(black)("mL"))) * ("1 mole Pb"("NO"_3)_2)/(10^3color(red)(cancel(color(black)("mL")))) = "0.01159 moles Pb"("NO"_3)_2#

Now, the two reactants take part in the reaction in a #2:3# mole ratio, so use that to figure out if you're dealing with a limiting reagent.

#0.009836 color(red)(cancel(color(black)("moles FeCl"_3))) * ("3 moles Pb"("NO"_3)_2)/(2color(red)(cancel(color(black)("moles FeCl"_3))))#

# = "0.01475 moles Pb"("NO"_3)_2#

As you can see, you have fewer moles of lead(II) nitrate available

#"0.01159 moles available " < " 0.01475 moles needed"#

than the number of moles needed to ensure that all the moles of iron(III) chloride take part in the reaction #-># lead(II) nitrate will act as a limiting reagent.

Notice that the reaction consumes lead(II) nitrate and produces lead(II) chloride, the precipitate, in a #3:3# mole ratio, which means that the reaction will produce the same number of moles of lead(II) chloride as the number of moles of lead(II) nitrate it consumes.

Since lead(II) nitrate acts as the limiting reagent, it will be completely consumed by the reaction. Consequently, the reaction will produce #0.01159# moles of lead(II) chloride.

To convert this to grams, use the compound's molar mass

#0.01159 color(red)(cancel(color(Black)("moles PbCl"_2))) * "278.1 g"/(1 color(red)(cancel(color(black)("mole PbCl"_2)))) = color(darkgreen)(ul(color(black)("3.22 g")))#

The answer is rounded to three sig figs.