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]

1 Answer
Oct 7, 2017

Answer:

#"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.