1oh shit! yes my mistake...
.... yup i overlooked dat moles of fe0 = moles of fe3o4
.....dats was a really nice question.... :)
thanx...
A 5.00 g sample containing Fe3O4, Fe2O3 and an inert impure substance, is treated with excess KI solution in presence of dil H2SO4. The entire iron is converted into Fe2+ along with liberation of iodine. The resulting solution is diluted to 50 mL. A 10 mL of the diluted solution requires 11.0 mL of 0.5 M Na2S2O3 solution to reduce the iodine present. A 25 mL of the diluted solution, after complete extraction of the iodine requires 6.40 ml of 0.50 M KMnO4 solution in dilute H2SO4 medium for the oxidation of Fe2+. Calculate the percentage of Fe2O3 and Fe3O4 in the original sample.
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5 Answers
1fe3o4= fe2o3 and feo
now fe3+ -----> fe2+ Iodine
no. of eq. of iodine = no. of eq. of fe3+ = 11X0.5X50/1000X10
=0.0275 (fe3+ from fe2o3 n fe3o4)
no. of eq. of fe2+ (this is during oxidation of fe2+ to fe3+) = 6.4X0.5X5X50/1000X25 = 0.032 (this is total fe2+)
so no. eq of fe3o4
=no of eq of feo (since fe3o4 = fe2o3 + fe0)
=0.032 - 0.0275
=0.0045
mol wt of fe3o4= 232
so wt of fe3o4 = 1.044gm.
% = 1.044/5 X100 =20.88 %
now, total eq. = 0.032
eq of fe3o4 = 0.0045
so total eq of fe2o3 = 0.0275 (fe2o3 from fe2o3 n fe3o4)
so actual fe2o3 eq = 0.0275/2 = 0.01375
mol wt of fe203= 160
so wt of fe2o3 =2.2gm
% = 2.2/5 X100 = 44%
{i have biiiiiiiiiiig doubts if at all this is correct !! :( :( :( }
1357Your first answer is correct
but the second one is wrong
the answer for second one is 29.6%
1357so no. eq of fe3o4
=no of eq of feo (since fe3o4 = fe2o3 + fe0)
=0.032 - 0.0275
=0.0045
here you have done a mistake because their moles are equal rather than equivalents
no. of eq of feo=0.0045=no. of moles of feo(=fe3o4)
no. of equivalents of fe2o3(from fe3o4)=0.0045*2=0.009
equivalents of fe2o3(from fe2o3)=0.0275-0.009=0.0185
Wt. =(0.0185/2)*160
%=wt/5=29.6%
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