Mole ConceptMedium
Question
The percentage of Fe(III) present in iron ore Fe0.93 O1.00 is (Fe = 56)
Options
A.94
B.6
C.21.5
D.15
Solution
$x \times ( + 3) + (0.93 - x) \times ( + 2) + 1 \times ( - 2) = 0 \Rightarrow x = 0.14$
Percentage of iron as Fe(III) = $\frac{0.14}{0.93} \times 100 = 15.05\%$
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