Ferrous ion was dissolved and diluted to250.0mL.25.00mL aliquot of the ferrous ion solution was titrated with0.01000M KMnO4 solution, and the mean of 3 acceptable, corrected titrations volumes was13.84mL.Calculate the mass of Fe in the250.0mL solution?

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Answer 1 · 2018-01-29T14:06:49

0.3864 g

Start with the 1/2 equations:

#sf(Fe^(2+)rarrFe^(3+)+e" " " "color(red)((1)))#

#sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((2)))#

To get the electrons to balance we X #sf(color(red)((1)))# by 5 and add to #sf(color(red)((2))rArr)#

#sf(5Fe^(2+)+MnO_4^(-)+8H^(+)+cancel(5e)rarr5Fe^(2+)+cancel(5e)+Mn^(2+)+4H_2O)#

#:.# 5 mol #sf(Fe^(2+)-=)# 1 mol #sf(MnO_4^(-))#

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(n_(MnO_4^(-))=0.01000xx13.84/1000=1.384xx10^(-4))#

#:.##sf(n_(Fe^(2+))=1.384xx10^(-4)xx5=6.920xx10^(-4))#

#sf(m=nxxA_r)#

#:.# The mass of iron(II) in 25.00 ml will be:

#sf(m_(Fe^(2+))=6.920xx10^(-4)xx55.85=0.03864color(white)(x)g)#

#:.# The mass present in the original 250.0 ml will be :

#sf(0.03864xx10=0.3864color(white)(x)g)#