# Question a9f4f

Apr 22, 2016

Here's what I got.

#### Explanation:

Since you didn't provide the concentration of the lead(II) nitrate solution, we're going to have to improvise a bit here.

You're dealing with a double replacement reaction in which aqueous potassium chromate, ${\text{K"_2"CrO}}_{4}$, and aqueous lead(II) nitrate, "Pb"("NO"_3)_2, will react to form the insoluble lead(II) chromate, ${\text{PbCrO}}_{4}$, and aqueous potassium nitrate, ${\text{KNO}}_{3}$.

The balanced chemical equation that describes this reaction looks like this

${\text{K"_ 2"CrO"_ (4(aq)) + "Pb"("NO"_ 3)_ (2(aq)) -> "PbCrO"_ (4(s)) darr + 2"KNO}}_{3 \left(a q\right)}$

Notice that the two reactants react in a $1 : 1$ mole ratio. This tells you that the reaction will always* consume equal numbers of moles** of potassium chromate and of lead(II) nitrate.

Moreover, lead(II) chromate is produced in the same $1 : 1$ mole ratio with both reactants, which means that the number of moles of lead(II) chromate produced by the reaction will match the number of moles of the reactants that take part in the reaction.

Use the molarity and volume of the potassium chromate solution to find how many moles of this reactant you have present

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

n_(K_2CrO_4) = "0.40 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

${n}_{{K}_{2} C r {O}_{4}} = {\text{0.00600 moles K"_2"CrO}}_{4}$

Now, you have three possible scenarios to account for here

$\textcolor{w h i t e}{a}$

• If you have fewer moles of lead(II) nitrate than moles of potassium chromate

In this case, lead(II) nitrate will be the limiting reagent because it will be completely consumed by the reaction before all the moles of potassium chromate will get the chance to react.

${n}_{P b {\left(N {O}_{3}\right)}_{2}} < \text{0.00600 moles}$

Since you have equal volumes of both solutions, you can say that if the concentration of the lead(II) nitrate is lower than that of potassium chromate, the former will act as a limiting reagent.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{P b {\left(N {O}_{3}\right)}_{2}} \textcolor{red}{<} {\text{0.40 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ lead(II) nitrate will be the limiting reagent

$\textcolor{w h i t e}{a}$

• if you have equal numbers of moles of each reactant

In this case, both of the reactants will be completely consumed by the reaction, i.e. you won't be dealing with a limiting reagent.

Since the lead(II) chromate is produced in a $1 : 1$ mole ratio with the two reactants, it follows that the reaction will produce $0.00600$ moles of ${\text{K"_2"CrO}}_{4}$.

Use the compound's molar mass to find how many Grams would contain this many moles

$0.00600 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles K"_2"CrO"_4))) * "184.19 g"/(1color(red)(cancel(color(black)("mole K"_2"CrO"_4)))) = color(green)(|bar(ul(color(white)(a/a)"1.1 g } \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{w h i t e}{a}$

• if you have more moles of lead(II) nitrate than moles of potassium chromate

This time, the potassium chromate will be completely consumed by the reaction before all the moles of lead(II) nitrate get the chance to react.

Potassium chromate will act as a limiting reagent. The reaction will once again produce $0.00600$ moles of lead(II) chromate. So, for

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{P b {\left(N {O}_{3}\right)}_{2}} \textcolor{red}{>} {\text{0.40 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ potassium chromate will be the limiting reagent

You will have

$0.00600 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles K"_2"CrO"_4))) * "184.19 g"/(1color(red)(cancel(color(black)("mole K"_2"CrO"_4)))) = color(green)(|bar(ul(color(white)(a/a)"1.1 g } \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answers are rounded to two sig figs.

It's worth noting that lead(II) chromate is a yellow insoluble solid that precipitates out of solution as the reaction proceeds.