$F{{e}_{2}}{{O}_{3}}+6HCl\to 2FeC{{l}_{3}}+3{{H}_{2}}O$

0,02$\to $0,12                                  (mol)

$CuO+2HCl\to CuC{{l}_{2}}+{{H}_{2}}O$

0,03$\to $0,06                                            (mol)

$\begin{align}  & {{n}_{F{{e}_{2}}{{O}_{3}}}}=\frac{{{m}_{F{{e}_{2}}{{O}_{3}}}}}{{{M}_{F{{e}_{2}}{{O}_{3}}}}}=\frac{3,2}{160}=0,02\left( mol \right) \\ & {{n}_{CuO}}=\frac{{{m}_{CuO}}}{{{M}_{CuO}}}=\frac{2,4}{80}=0,03\left( mol \right) \\\end{align}$

${{n}_{HCl}}=6.{{n}_{F{{e}_{2}}{{O}_{3}}}}+2.{{n}_{CuO}}=6.0,02+2.0,03=0,18\left( mol \right)$

${{V}_{HCl}}=\frac{{{n}_{HCl}}}{{{C}_{{{M}_{HCL}}}}}=\frac{0,18}{0,2}=0,9(l)$

$\Rightarrow $Chọn đáp án D