Question

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]

Answer 1

`"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, `"FeSO"_4`.

So tetraoxosulfate(VI) is an alternative name used for the sulfate anion, `"SO"_4^(2-)`. The name suggests the number of oxygen atoms present in the anion

`"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

`"Fe"_ ((aq))^(2+) + 2"e"^(-) -> "Fe"_ ((s))`

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

`0.03684 color(red)(cancel(color(black)("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.