Redox and Equivalent ConceptMedium
Question
An oxide of iron contains 30% oxygen by mass. The oxidation state of iron in this oxide is (Fe =56)
Options
A.+1
B.+2
C.+3
D.+4
Solution
$\frac{N_{Fe}}{N_{O}} = \frac{70/56}{30/16} = \frac{2}{3} \Rightarrow \text{Formula = F}\text{e}_{2}O_{3}$
∴ Oxidation state of Fe = +3.
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