Đáp án đúng: D
6,4
$\sum{{{n}_{BaC{{O}_{3}}}}}=\,\frac{5,91\,+\,3,94}{197}\,=\,0,05\,\,mol\,\,\Rightarrow \,{{n}_{C{{O}_{2}}}}\,=\,0,05\,mol$
${{n}_{CO}}\,=\,{{n}_{C{{O}_{2}}}}\,=\,0,05\,mol$
$\begin{array}{l}\xrightarrow{BTKL}\,{{m}_{F{{e}_{2}}{{O}_{3}}}}\,+\,{{m}_{CO}}\,=\,{{m}_{X}}\,+\,{{m}_{C{{O}_{2}}}}\\\Rightarrow \,{{m}_{X}}\,=\,7,2\,+\,0,05.28\,-\,0,05.44\,=\,6,4\,gam\end{array}$