Question

In puec A fe 0 , are in ratio 1: 1 Whe no of Fe"t are = 0.93 Missing ( f e^{2} t=0.07 )
sunce calh fe ( ^{2} ) has 2 fue chage, total due to mistug ( 0.07=0.07 times 2=0.14 ) chacye due to whryvuy To mantain newtrality ( , 0.14 ) twe chasge is compensaled by fe 3 tom henlacing one fe com by one fe "con incecases one twe charge ( 0 cdot 14 ) twe changes must be compensaled by 0.14 fe ist ions
1. ( 0.93 mathrm{fe}^{2 mathrm{f}} ) iom have ( 0.14 mathrm{fe}^{3+} ) com Iftersa loo fe ( ^{2+} ) iom have ( =frac{0.14}{0.93} times 100 )
[
operatorname{Aus}=frac{15.05 %}{(B)}
]

# The composition of a sample of Wustite is Fe0.930. What is the percentage of iron present as Fet in total iron ? A 25.05% B 15.05% 35.05% 45.05%

Solution