Đáp án đúng: A
33,6 gam
$\left\{ \begin{array}{l}{{n}_{CO}}\,+\,{{n}_{C{{O}_{2}}}}\,=\,\frac{35,84}{22,4}\,=\,1,6\,mol\\28{{n}_{CO}}\,+\,44{{n}_{C{{O}_{2}}}}\,=\,1,6.18.2\,=\,57,6\,gam\end{array} \right.$
$\Rightarrow \,{{n}_{C{{O}_{2}}}}\,=\,{{n}_{CO}}\,=\,0,8\,mol$
$\displaystyle \Rightarrow {{n}_{Fe}}\,=\,\frac{3}{4}{{n}_{O\,\,(\text{ox}it)}}\,=\,\frac{3}{4}{{n}_{C{{O}_{2}}}}\,\Rightarrow \,{{m}_{Fe}}\,=\,33,6\,gam$