# How many moles of electrons are required to deposit 5.6g of Iron from a solution of Iron(II) tetraoxosulphate(VI) ? [Fe = 56, S = 32, O = 16]

Oct 7, 2017

${\text{0.074 moles e}}^{-}$

#### Explanation:

For starters, it's worth pointing out that iron(II) tetraoxosulfate(VI) is just a fancy name for iron(II) sulfate, ${\text{FeSO}}_{4}$.

So tetraoxosulfate(VI) is an alternative name used for the sulfate anion, ${\text{SO}}_{4}^{2 -}$. The name suggests the number of oxygen atoms present in the anion

$\text{tetraoxo = 4 oxygen atoms}$

and the oxidation state of sulfur in this anion

("VI") = +"6 oxidation state for S"

Now, before doing anything else, use the molar mass of iron(II) sulfate to calculate the number of moles of this compound present in your sample.

5.6 color(red)(cancel(color(black)("g"))) * "1 mole FeSO"_4/(152color(red)(cancel(color(black)("g")))) = "0.03684 moles FeSO"_4

Since $1$ mole of iron(II) sulfate contains $1$ mole of iron(II) cations, you can say that your solution will contain $0.03684$ moles of iron(II) cations.

In order to reduce the iron(II) cations to iron metal, you need

${\text{Fe"_ ((aq))^(2+) + 2"e"^(-) -> "Fe}}_{\left(s\right)}$

So $1$ mole of iron(II) cations requires $2$ moles of electrons, which means that your sample will require

$0.03684 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles Fe"^(2+)))) * "2 moles e"^(-)/(1color(red)(cancel(color(black)("mole Fe"^(2+))))) = color(darkgreen)(ul(color(black)("0.074 moles e}}^{-}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of iron(II) sulfate.