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
Explanation:
For starters, it's worth pointing out that iron(II) tetraoxosulfate(VI) is just a fancy name for iron(II) sulfate,
So tetraoxosulfate(VI) is an alternative name used for the sulfate 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
In order to reduce the iron(II) cations to iron metal, you need
#"Fe"_ ((aq))^(2+) + 2"e"^(-) -> "Fe"_ ((s))#
So
#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.