Question

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

Answer 1

`"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`