What fraction of Fe exists as Fe (III) in Fe _ 0.96 O ? (Consider Fe _ 0.96 to be made up of Fe (II) and Fe (III) only)

Step-by-step solution

Let's assume Fe in Fe ²⁺= x Let's assume Fe in Fe ³⁺(0.96-x) Using the charge conservation formula for Fe _ 0.96 O _ 1.00 We get: (x)(+2)+(0.96-x)(+3)-2=0 aligned & =2 x+2.88-3 x-2=0 & x=0.88 aligned So, number of ions of Fe ³⁺=0.96-x=0.96-0.88=0.08 Fraction of Fe ³⁺ ion = 0.08 0.96 = 1 12