Đáp án đúng: C
4,04.
$\begin{array}{l}{{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{18,56}{232}=0,08\,mol\\F{{e}_{3}}{{O}_{4}}+8HCl\xrightarrow{{}}2FeC{{l}_{3}}+FeC{{l}_{2}}+4{{H}_{2}}O\\0,075\,\leftarrow \,0,6\,\xrightarrow{{}}\,\,\,\,\,\,0,15\,\,\to \,\,\,0,075\end{array}$
$\begin{array}{l}\,\,\,\,Cu\,\,\,\,\,+\,\,\,\,2F{{e}^{3+}}\xrightarrow{{}}C{{u}^{2+}}+2F{{e}^{2+}}\\0,075\leftarrow \,0,15\end{array}$
$x=(0,08-0,075).232+(0,12-0,075).64=4,04\,gam$