Đáp án đúng: B
a = b.
$\begin{array}{l}4FeC{{O}_{3}}+{{O}_{2}}\xrightarrow{{{t}^{o}}}2F{{e}_{2}}{{O}_{3}}+4C{{O}_{2}}\\\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{a}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\end{array}$
$\begin{array}{l}4Fe{{S}_{2}}+11{{O}_{2}}\xrightarrow{{{t}^{o}}}2F{{e}_{2}}{{O}_{3}}+8S{{O}_{2}}\\\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{11b}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2b\end{array}$
Do${{P}_{trc}}={{P}_{sau}}$ ⇒số mol khí trước và sau phản ứng không thay đổi.
Ta có:${{n}_{{{O}_{2}}\,pu}}={{n}_{C{{O}_{2}}}}+{{n}_{S{{O}_{2}}}}\Rightarrow \frac{a}{4}+\frac{11b}{4}=a+2b$
$\displaystyle \Rightarrow a\text{ }+\text{ }11b\text{ }=\text{ }4a\text{ }+\text{ }8b~\Rightarrow ~a\text{ }=\text{ }b.$