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

1 Answer
Apr 24, 2017

Answer:

Approx. #76.5*mL#.........

Explanation:

#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.