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the composition of wustite is Fe(0.93)O.what % of iron is present in the form of Fe ||

Asked by rishita gajbhiye 14th February 2011, 11:19 AM
Answered by Expert
Answer:
Dear Student
 
Given is Fe0.93O1.00
 
let the no. of Fe2+=x
no. of Fe3+=0.93-x
As we know the total charge must be zero fo a neutral molecule.
 
2x + (0.93-x)3 - 2 = 0
2x + 2.79 - 3x -2 = 0
 
solving, x=0.79
thus % of Fe(II)=79%
 
We hope that clarifies your query.
Regards
Team
Topperlearning
Answered by Expert 15th February 2011, 11:34 AM
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