Xét quá trình cho – nhận e:

Cho e:

\(\begin{align} & \text{Fe: }\overset{\text{0}}{\mathop{\text{Fe}}}\,\to \overset{\text{+3}}{\mathop{\text{Fe}}}\,\text{ + 3e} \\ & \text{FeO:}\overset{\text{+2}}{\mathop{\text{ Fe}}}\,\to \overset{\text{+3}}{\mathop{\text{Fe}}}\,\text{+}\text{1e} \\ & \text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{:}\overset{\text{+}\frac{\text{8}}{\text{3}}}{\mathop{\text{ F}{{\text{e}}_{\text{3}}}}}\,\to \text{3}\overset{\text{+3}}{\mathop{\text{Fe}}}\,\text{ + 1e} \\ \end{align}\)

Nhận e: 

\(\overset{+6}{\mathop{S}}\,\to \overset{+4}{\mathop{S}}\,+2e\)

A. Fe, Fe₂O₃: \({{n}_{S{{O}_{2}}}}=\frac{3}{2}=1,5mol\)

B. FeO, FeO: \({{n}_{S{{O}_{2}}}}=\frac{13+1}{2}=2\text{ }mol\)

C. Fe₃O₄, Fe₂O₃: \({{n}_{S{{O}_{2}}}}=\frac{1}{2}=\text{ }0,5\text{ }mol\)

D. FeO, Fe₃O₄: \({{n}_{S{{O}_{2}}}}=\frac{1+1}{2}=1\text{ }mol\)

⇒ Chọn D