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  • In the compound , the percent of iron existing as Fe(III) is:<div

In the compound \small Fe_{0.85}O, the percent of iron existing as Fe(III) is:

Option: 1

25.3


Option: 2

30.3


Option: 3

35.3


Option: 4

40.3


Answers (1)

best_answer

For every 1 mole of the oxygen atom, 0.85 moles of iron atoms are present. If r is the amount of Fe(III) present, then the electrical neutrality requires that

x\times3+(0.85-x)\times2=1\times2
This gives  x = 0.3 mol


Percent of Fe(III) =  \frac{0.3}{0.85}\times100%

                          = 35.3%

 

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mansi

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