# Question 1aa1a

Apr 23, 2016

${\text{5.6 g PbCl}}_{2}$

#### Explanation:

The key to this problem is the balanced chemical equation that describes this double replacement reaction.

Aqueous lead(II) nitrate, "Pb"("NO"_3)_2, and aqueous magnesium chloride, ${\text{MgCl}}_{2}$, will react to form lead(II) chloride, an insoluble solid that precipitates out of solution, and aqueous magnesium nitrate, "Mg"("NO"_3)_2.

"Pb"("NO"_ 3)_ (2(aq)) + "MgCl"_ (2(aq)) -> "PbCl"_ (2(s)) darr + "Mg"("NO"_ 3)_ (2(aq))

The two reactants are consumed by the reaction in a $1 : 1$ mole ratio, so right from the start you know that the reactant that has the fewest moles present will act as a limiting reagent, i.e. it will be completely consumed by the reaction.

Moreover, lead(II) chloride is produced in a $1 : 1$ mole ratio with both reactants, which means that the reactant that acts as a limiting reagent will determine how many moles, and implicitly grams, of lead(II) chloride will be produced.

So, use the molarities and volumes of the two solutions to find the number of moles of each reactant

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

n_(Pb(NO_3)_2) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(200 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

n_(Pb(NO_3)_2) = "0.020 moles Pb"("NO"_3)_2

and

n_(MgCl_2) = "0.200 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(300 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{M g C {l}_{2}} = {\text{0.060 moles MgCl}}_{2}$

Notice that you have fewer moles of lead(II) nitrate than of magnesium chloride, which means that the lead(II) nitrate will be completely consumed by the reaction.

This implies that the reaction will produce

0.020 color(red)(cancel(color(black)("moles Pb"("NO"_3)_2))) * "1 mole PbCl"_2/(1color(red)(cancel(color(black)("mole Pb"("NO"_3)_2)))) = "0.020 moles PbCl"_2#

To get the mass of lead(II) chloride that will precipitate out of solution, use the compound's molar mass

$0.020 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles PbCl"_2))) * "278.11 g"/(1color(red)(cancel(color(black)("mole PbCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)"5.6 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.

SIDE NOTE Lead(II) chloride is a white precipitate. 