Chemistry - Redox Reactions Question with Solution | TestHub

Let mmoles of each be $x$.

The n-factor of FeO is 1.

The n-factor of Fe₀.₈₀O is 0.4.

The milliequivalents (meq) of FeO + meq of Fe₀.₈₀O = meq of KMnO₄.

$$x\times 1+x\times 0.4=35\times 0.3\times 5$$

$$1.4x=52.5$$

$$x=\frac{52.5}{1.4}$$

$$x=37.50mmoles$$

mmoles of Fe³⁺ produced = $37.5+37.5\times 0.8=37.5+30=67.50mmoles$.