Đáp án đúng: A
50% và 50%
$\left\{ \begin{array}{l}{{n}_{C{{O}_{2}}}}\,+\,{{n}_{CO}}\,=\,0,03\,mol\\44{{n}_{C{{O}_{2}}}}\,+\,28{{n}_{CO}}\,=\,0,03.19,33.2\,=\,1,1598\end{array} \right.$$\Rightarrow \left\{ \begin{array}{l}{{n}_{C{{O}_{2}}}}\,=\,0,02\\{{n}_{CO}}\,=\,0,01\end{array} \right.$
$\left\{ \begin{array}{l}80{{n}_{CuO}}\,+\,160{{n}_{F{{e}_{2}}{{O}_{3}}}}\,=\,3,2\\{{n}_{CuO}}\,+\,3{{n}_{F{{e}_{2}}{{O}_{3}}}}\,=\,2{{n}_{C{{O}_{2}}}}\,+\,{{n}_{CO}}\,=\,0,05\end{array} \right.$$\Rightarrow \,\left\{ \begin{array}{l}{{n}_{CuO}}\,=\,0,02\\{{n}_{F{{e}_{2}}{{O}_{3}}}}\,=\,0,01\end{array} \right.\,$
$\Rightarrow {{m}_{CuO}}\,=\,1,6\,gam\,\Rightarrow \,{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}{{m}_{CuO}}\,=\,50{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}$