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

Jan 8, 2017

$\text{5Fe"^(2+) + "MnO"_4^(-) + 8"H"^+ → "5Fe"^(3+) + "Mn"^(2+) + 4"H"_2"O}$

#### Explanation:

The stoichiometric equation is as given, and shows that 5 equiv of $F {e}^{2 +}$ will react with 1 equiv of permanganate in acidic media.

Because ${\text{Mn}}^{2 +}$ 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.

$\text{Moles of permanganate} = 10 \times {10}^{-} 3 L \times 0.010 \cdot m o l \cdot {L}^{-} 1 = 1.0 \times {10}^{-} 4 \cdot m o l .$

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 $\text{iron(II) sulfate heptahydrate}$

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