What volume of nitric acid of $5.0molL^-1$ concentration is required to oxidize a $16*g$ mass of $Fe^(2+)$ ions?

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Answer 1 · 2017-04-24T17:22:51

Approx. $76.5*mL$ .........

$Fe^(2+)$ is oxidized to $"ferric ion"$ :

$"Fe"^(2+) rarr "Fe"^(3+) + e^-$ $(i)$

And nitric acid is reduced to nitrous oxide:

$"H"stackrel(V+)"NO"_3 + "3H"^(+) + 3e^(-) rarrstackrel(II+)"NO" +"2H"_2"O"$ $(ii)$

And so $3xx(i)+(ii)$ gives............

$3Fe^(2+) +4HNO_3 rarr 3Fe^(3+) +NO(g)uarr +3NO_3^(-) +2H_2O(l)$

I have treated $H^(+)-=HNO_3$ here...........

Now you have claimed you got $16*g$ of ferrous iron (which is less than the equivalent mass of ferrous salt, like $FeSO_4$ .

Going with $16*g$ , this represents a molar quantity of........

$(16*g)/(55.8*g*mol^-1)=0.287*mol$ , and thus we need $4/3xx0.287*mol=0.382*mol$ with respect to nitric acid.

And finally, $"moles"="concentration"xx"volume"$ , or $"volume"="moles"/"concentration"$ ,

$=(0.382*mol)/(5.0*mol*L^-1)=76.5*mL$ .......

Remember I have calculated with respect to a mass of ferrous iron, not to a mass of ferrous salt.

What volume of nitric acid of 5.0*mol*L^-15.0*mol*L^-1 concentration is required to oxidize a 16*g16*g mass of Fe^(2+)Fe^(2+) ions?