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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: The percentage of
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juana (44)

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Fe3+ in a sample of Fe0.93 O1.0 is ?


 

    
juana (44)

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oops sorry for the repeated post of the question.
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rohitkuruvila (311)

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is it approx 76.5%

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aditi_g (355)

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the method is...
let number ofFe2+ be x and number of Fe3+ be y
so x+ y=93
balancing charge with that of oxygen we get
2x + 3y = 200
solving these 2 eqns we get y= 14
so %age = 14/93 *100= 15 % approx
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juana (44)

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rohit it's 15.05
what method did you use to get 76.5%?
If you want the solutions have a look at the other post of this question. Some ppl have answered there.
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sahana_2193 (26)

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its 75% in wustite.
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