Đáp án đúng: A
116,89.
Qui hỗn hợp đầu về: Fe2O3; x mol FeO; y mol Cu
X + HCl dư và không có kết tủa sau đó => Cu phản ứng hết
$\displaystyle \text{2FeC}{{\text{l}}_{\text{3}}}\text{+Cu}\xrightarrow{{}}\text{2FeC}{{\text{l}}_{\text{2}}}\text{+CuC}{{\text{l}}_{\text{2}}}$
Y gồm 0,08 mol FeCl3; (x+2y) mol FeCl2; y mol CuCl2; HCl
⇒(0,08.3+2x+2y) < 0,9
Hỗn hợp đầu gồm: (0,04+y) mol Fe2O3; x mol FeO; y mol Cu
$\displaystyle \Rightarrow \text{27,2=160}\text{.}\left( \text{0,04+y} \right)\text{+72x+64y}\,\,{{\,}^{\left( \text{1} \right)}}$
Khi điện phân:
Catot(-): thứ tự có thể xảy ra
$\displaystyle \text{F}{{\text{e}}^{\text{3+}}}\text{+1e}\xrightarrow{{}}\text{F}{{\text{e}}^{\text{2+}}}$
$\displaystyle \text{C}{{\text{u}}^{\text{2+}}}\text{+2e}\xrightarrow{{}}\text{Cu}$
$\displaystyle \text{2}{{\text{H}}^{\text{+}}}\text{+2e}\xrightarrow{{}}{{\text{H}}_{\text{2}}}\left( \text{*} \right)$
$\displaystyle \text{F}{{\text{e}}^{\text{2+}}}\text{+2e}\xrightarrow{{}}\text{Fe}$
Anot(+):
$\displaystyle \text{2C}{{\text{l}}^{\text{-}}}\xrightarrow{{}}\text{C}{{\text{l}}_{\text{2}}}\text{+2e}$
Vì ngừng điện phân khi catot có khí => dừng trước quá trình (*)
Bảo toàn e:$\displaystyle {{\text{n}}_{\text{C}{{\text{l}}_{\text{2}}}}}\text{=}\frac{\text{1}}{\text{2}}\left( \text{0,08+2y} \right)\text{=0,04+y}\,\left( \text{mol} \right)$
=> mgiảm$\displaystyle \text{=}{{\text{m}}_{\text{Cu}}}\text{+}{{\text{m}}_{\text{C}{{\text{l}}_{\text{2}}}}}\text{=y}\text{.64+}\left( \text{0,04+y} \right)\text{.71=13,64g}$
$\displaystyle \Rightarrow \text{y=0,08}\,\text{mol}$. Từ$\displaystyle \left( \text{1} \right)\Rightarrow \text{x=0,04}\,\text{mol}$
=> Sau điện phân còn: nHCl dư = 0,1 mol;$\displaystyle {{\text{n}}_{\text{FeC}{{\text{l}}_{\text{2}}}}}\text{=0,16+0,04+0,08=0,28}\,\text{mol}$
$\displaystyle \begin{array}{l}\,\,\text{3F}{{\text{e}}^{\text{2+}}}\,\,\text{+}\,\,\,\,\,\text{4}{{\text{H}}^{\text{+}}}\,\,\text{+NO}_{\text{3}}^{\text{-}}\xrightarrow{{}}\text{3F}{{\text{e}}^{\text{3+}}}\text{+NO+2}{{\text{H}}_{\text{2}}}\text{O}\\\text{0,075}\leftarrow \text{0,1 mol}\end{array}$
$\displaystyle \text{F}{{\text{e}}^{\text{2+}}}\text{+A}{{\text{g}}^{\text{+}}}\xrightarrow{{}}\text{F}{{\text{e}}^{\text{3+}}}\text{+Ag}$
$\displaystyle \text{A}{{\text{g}}^{\text{+}}}\text{+C}{{\text{l}}^{\text{-}}}\xrightarrow{{}}\text{AgCl}$
=> Kết tủa gồm: 0,205 mol Ag; 0,66 mol AgCl
$\displaystyle \Rightarrow \text{m=116,85}\,\text{gam}$
