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
Explanation:
You're dealing with the oxidation of the iron(II) cations,
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
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)|)))