What mass of iron(ll) sulfate heptahydrate would completely react with approximately 10 ml of 0.010 M #KMnO_4#?

1 Answer
anor277
Jan 8, 2017

#"5Fe"^(2+) + "MnO"_4^(-) + 8"H"^+ → "5Fe"^(3+) + "Mn"^(2+) + 4"H"_2"O"#

Explanation:

The stoichiometric equation is as given, and shows that 5 equiv of #Fe^(2+)# will react with 1 equiv of permanganate in acidic media.

Because #"Mn"^(2+)# is almost completely colourless, the endpoint is signalled by the persistence of the deep crimson colour of permanganate ion. This is an easy endpoint to vizualize.

#"Moles of permanganate"=10xx10^-3Lxx0.010*mol*L^-1=1.0xx10^-4*mol.#

Because of the molar equivalence given in the reaction, 5 equiv of ferrous ion will be oxidized. And so we mulitply this molar quantity by the formula mass of #"iron(II) sulfate heptahydrate"#

#5xx1.0xx10^-4*molxx278.02*"g"*"mol"^-1=??g#

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