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%
1
skygirl
·2008-11-07 10:01:01
why?? plz explain... i have thot much and found this one...
1357
Manish Shankar
·2008-11-07 10:55:26
so 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%
1
skygirl
·2008-11-07 17:22:20
oh shit! yes my mistake...
.... yup i overlooked dat moles of fe0 = moles of fe3o4
.....dats was a really nice question.... :)
thanx...
