What volume of 0.825 M #MnO_4 ^-# is needed to completely react with 50.0 mL of 0.520 M #Fe(II)#?

1 Answer
Jun 27, 2016

Answer:

#"158 mL"#

Explanation:

You're dealing with the oxidation of the iron(II) cations, #"FE"^(2+)#, to iron(III) cations, #"Fe"^(3+)#, by the permanganate anion, #"MnO"_4^(-)#, which gets reduced to manganese(II) cations, #"Mn"^(2+)#.

Your starting point here will thus be to write a balanced chemical equation for this redox reaction.

Now, I will not show you how to balance this equation here because that would make for a very long answer and because that would take the focus from the actual question.

#color(red)(5)"Fe"_ ((aq))^(2+) + "MnO"_ (4(aq))^(-) + 8"H"_ ((aq))^(+) -> 5"Fe"_ ((aq))^(3+) + "Mn"_ ((aq))^(2+) + 4"H"_ 2"O"_ ((l))#

Notice that the reaction requires #1# mole of permanganate anions for every #color(red)(5)# moles of iron(II) cations present in solution.

Your goal now is to use the molarity and volume of the iron(II) solution to determine how many moles of iron(II) cations it contained

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))#

Plug in your values to find

#n_("Fe"^(2+)) = "0.520 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(50.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#n_("Fe"^(2+)) = "0.0260 moles Fe"^(2+)#

According to the balanced chemical equation, this many moles of iron(II) cations will require

#0.0260 color(red)(cancel(color(black)("moles Fe"^(2+)))) * (color(red)(5)color(white)(a)"moles MnO"_4^(-))/(1color(red)(cancel(color(black)("mole Fe"^(2+))))) = "0.130 moles MnO"_4^(-)#

Since you know the molarity of the permanganate solution, all you have to do now is figure out what volume would contain that many moles of permanganate anions

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c color(white)(a/a)|)))#

Plug in your values to find

#V_("MnO"_4^(-)) = (0.130 color(red)(cancel(color(black)("moles"))))/(0.825color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.15758 L"#

Rounded to three sig figs and expressed in milliliters, the answer will be

#V_("MnO"_4^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)("158 mL")color(white)(a/a)|)))#