Identify the correct relationships regarding the n-factor and equivalents of (non-stoichiometric wus — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

1 mole of $Fe_{0.95}O$ contains 0.95 moles of Fe. Initial average O.S. = +200/95. Final O.S. = +3. Total change for 0.95 atoms = $(3 - 200/95) \times 0.95 = 0.85$. Thus n=0.85 and $E = M/0.85$. The percentage of $Fe^{3+}$ is calculated via $3y + 2(0.95-y) = 2 \Rightarrow y = 0.1$, which is $0.1/0.95 = 10.5\%$. D is false ($M/0.85 = 20M/17$).