A compound Fe0.93O1.00, present find out the % of Fe2+ and Fe3+.
Note: Pls give the solution.....
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We have Fe3+=x and Fe2+=.93-x
then using charge balance..
3*x+2*(0.93-x)=2*1
( 1 for oxygen)
x=.14
% of Fe3+=(.14/.93)*100=15%
% of Fe2+=(.79/.93)*100=85%
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