# Question e11f9

Apr 25, 2016

$\text{97.2% Fe}$

#### Explanation:

The idea here is that you need to backtrack from the balanced chemical reaction that describes the oxidation of iron(II) cations to iron(III) cations in an acidified solution of potassium dichromate, ${\text{K"_2"Cr"_2"O}}_{7}$, to find how many moles of iron(II) cations are present in the $\text{20 mL}$ sample.

Once you know that, you can find the number of moles of iron(II) cations in the $\text{250 mL}$ sample, and finally the mass of iron that was dissolved in the initial solution of sulfuric acid.

So, the balanced chemical equation that describes this redox reaction looks like this

$\textcolor{red}{6} {\text{FeSO"_ (4(aq)) + "K"_ 2"Cr"_ 2"O"_ (7(aq)) + 7"H"_ 2"SO"_ (4(aq)) -> "Cr"_ 2("SO"_ 4)_ (3(aq)) + 3"Fe"_ 2("SO"_ 4)_ (3(aq)) + 7"H"_ 2"O"_ ((l)) + "K"_ 2"SO}}_{4 \left(a q\right)}$

To keep things simple, I'll eliminate spectator ions and use the net ionic equation

$\textcolor{red}{6} {\text{Fe"_ ((aq))^(2+) + "Cr"_ 2"O"_ (7(aq))^(2-) + 14"H"_ ((aq))^(+) -> 2"Cr"_ ((aq))^(3+) + 6"Fe"_ ((aq))^(3+) + 7"H"_ 2"O}}_{\left(l\right)}$

So, the reaction consumes $1$ mole of dichromate anions, ${\text{Cr"_2"O}}_{7}^{2 -}$, for every $\textcolor{red}{6}$ moles of iron(II) cations present in solution.

Use the molarity and volume of the potassium dichromate solution, which dissociates completely in aqueous solution to form dichromate anions in a $1 : 1$ mole ratio, to find how many moles of dichromate anions were used in the titration.

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

You will have

n_(Cr_2O_7^(2-)) = "0.0209 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(18.1 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{C {r}_{2} {O}_{7}^{2 -}} = {\text{0.0003783 moles Cr"_2"O}}_{7}^{2 -}$

So, if this many moles of dichromate anions were needed, it follows that the solution contained

0.0003783 color(red)(cancel(color(black)("moles Cr"_2"O"_7^(2-)))) * (color(red)(6)color(white)(a)"moles Fe"^(2+))/(1color(red)(cancel(color(black)("mole Cr"_2"O"_7^(2-))))) = "0.002270 moles Fe"^(2+)

Now, this is how many moles of iron(II) cations are present in your $\text{20 mL}$ aliquot, which means that the $\text{250 mL}$ solution contained

250 color(red)(cancel(color(black)("mL solution"))) * "0.002270 moles Fe"^(2+)/(20color(red)(cancel(color(black)("mL solution")))) = "0.02838 moles Fe"^(2+)

Now, you don't need to write a balanced chemical equation that describes what happens when iron metal is dissolved in sulfuric acid to form iron(II) cations.

All you have to do now is assume that all the moles of iron metal that were present in the wire were converted to moles of iron(II) cations

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{moles of Fe"_ ((s)) -> " moles of Fe}}_{\left(a q\right)}^{2 +}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You can find the mass of iron present in the wire by using the element's molar mass

0.02838 color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = "1.5849 g"

This means that the percent composition of iron in that wire will be

"% Fe" = (1.5849 color(red)(cancel(color(black)("g"))))/(1.63color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"97.2% Fe"color(white)(a/a)|)))#

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