Đáp án đúng: D
110,20g
$\displaystyle \underbrace{F{{e}_{2}}{{O}_{3}},\,Cu}_{\text{hh H}}\xrightarrow{HCl}\left\langle \begin{array}{l}Cu\,(k.\tan )\\\underbrace{F{{e}^{2+}},\,C{{u}^{2+}},\,{{H}^{+}}(1),\,C{{l}^{-}}}_{\text{40,36}\,\text{(g)}\,\text{c}\text{.tan}\,}\xrightarrow{AgN{{O}_{3}}}\underbrace{F{{e}^{3+}},\,C{{u}^{2+}},\,N{{O}_{3}}^{-}}_{dd}+\underbrace{NO}_{0,01\,mol}+\underbrace{Ag,\,AgCl}_{m\,(g)\,Z}\end{array} \right.$
Ta có :
$\begin{array}{l}{{n}_{HCl(1)}}=4{{n}_{NO}}=0,04\,mol\\\Rightarrow {{n}_{F{{e}_{2}}{{O}_{3}}}}={{n}_{Cu(pu)}}=\frac{{{m}_{\text{c}\text{.tan}}}-36,5{{n}_{HCl(1)}}}{2{{M}_{FeC{{l}_{2}}}}+{{M}_{CuC{{l}_{2}}}}}=\frac{40,36-0,04.36,5}{389}=0,1\,mol\end{array}$
$\begin{array}{l}\left\{ \begin{array}{l}\xrightarrow{BT:\,e}{{n}_{Ag}}={{n}_{FeC{{l}_{2}}}}-3{{n}_{NO}}=0,2-0,01.3=0,17\,mol\\\xrightarrow{BT:\,Cl}{{n}_{AgCl}}={{n}_{HCl}}=6{{n}_{F{{e}_{2}}{{O}_{3}}}}+{{n}_{HCl(1)}}=0,64\,mol\end{array} \right.\,\,\\\Rightarrow {{m}_{\downarrow }}=108{{n}_{Ag}}+143,5{{n}_{AgCl}}=110,2\,gam\end{array}$
