IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

juana

Fe³⁺in a sample of Fe_0.93 O_1.0 is ?

smarsh

let the no of fe(+2)be x then fe(+3) will be 93-x

then acc to conservation of charge:-

2x+(93-x)3=200
which givesx=79 ....thus fe(+3)is 14 and the percentage is
(14/93)*100

that comes out to be round abt 15.05

atul_sinha89

LET THE NUMBER OF Fe³⁺ IONS BE X THEN, NO. OF Fe²⁺ IONS= 0.93-X

NOW ,

OXIDATION STATE OF Fe IS +2.

=> 3X+2(0.93-X) = 2

=> 3X+1.86-2X = 2

=>X = 2-1.86

=>X = 0.14

NUMBER OF ATOMS OF Fe³⁺ IONS IS 0.14

%AGE OF Fe³⁺ = (0.14/0.93)*100

=15.05%

RATE ME IS MY ANSWER IS SATISFACTORY..

kria

yupp answer is 15.05% ................! fe^3+

kria

wll if fe2+ is 93 present 4 every 100 oxide......

fe3+ is 93-x

due to electrical nuetrality we hav:-

2x+ (93-x)3=100*2

2x+279-3x=200

x=79

therefore fraction of fe2+= 79/93*100

=84.94%=85%

fe3+= 100-85=15%

hope i m right...........!

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