Đáp án đúng: D
86,9.
${{\text{n}}_{\text{e}}}\text{=}\frac{\text{It}}{\text{F}}\text{=0,6}\,\text{mol}$
Các quá trình có thể xảy ra là
Catot
Anot
$\begin{array}{l}\text{C}{{\text{u}}^{\text{2+}}}\text{+2e}\xrightarrow{{}}\text{Cu}\\\text{2}{{\text{H}}_{\text{2}}}\text{O+2e}\xrightarrow{{}}\text{2O}{{\text{H}}^{\text{-}}}\text{+}{{\text{H}}_{\text{2}}}\end{array}$
$\displaystyle \begin{array}{l}\text{2C}{{\text{l}}^{\text{-}}}\xrightarrow{{}}\text{C}{{\text{l}}_{\text{2}}}\text{+2e}\\\text{2}{{\text{H}}_{\text{2}}}\text{O}\xrightarrow{{}}\text{4}{{\text{H}}^{\text{+}}}\text{+}{{\text{O}}_{\text{2}}}\text{+4e}\end{array}$
$\left\{ \begin{array}{l}{{\text{n}}_{\text{C}{{\text{l}}_{\text{2}}}}}\text{+}{{\text{n}}_{{{\text{O}}_{\text{2}}}}}\text{=0,2}\,\text{mol}\\{{\text{n}}_{\text{e}}}\text{=2}{{\text{n}}_{\text{C}{{\text{l}}_{\text{2}}}}}\text{+4}{{\text{n}}_{{{\text{O}}_{\text{2}}}}}\text{=0,6}\,\text{mol}\end{array} \right.\Rightarrow {{\text{n}}_{\text{C}{{\text{l}}_{\text{2}}}}}\text{=}{{\text{n}}_{{{\text{O}}_{\text{2}}}}}\text{=0,1}\,\text{mol}\,\Rightarrow {{\text{n}}_{\text{NaCl}}}\text{=0,2}\,\text{mol}$
Vì khi thêm Fe vào thì tạo NO ⇒H+ dư.
Chất rắn thu được gồm 2 kim loại ⇒ tạo$\text{F}{{\text{e}}^{\text{2+}}}$,$\text{C}{{\text{u}}^{\text{2+}}}$ chuyển thành Cu và$\text{C}{{\text{u}}^{\text{2+}}}$còn dư sau điện phân.
⇒${{\text{n}}_{\text{C}{{\text{u}}^{\text{2+}}}\,\text{dp}}}\text{=0,3}\,\text{mol}$
Có các phản ứng sau
$\begin{array}{l}\text{3Fe+8}{{\text{H}}^{\text{+}}}\text{+NO}_{\text{3}}^{\text{-}}\xrightarrow{{}}\text{3F}{{\text{e}}^{\text{2+}}}\text{+2NO+4}{{\text{H}}_{\text{2}}}\text{O}\\\text{0,15}\,\,\,\text{0,4}\,\,\,\,\,\text{0,05}\end{array}$
$\begin{array}{l}\text{Fe+C}{{\text{u}}^{\text{2+}}}\xrightarrow{{}}\text{F}{{\text{e}}^{\text{2+}}}\text{+Cu}\\\,\text{x}\,\,\,\,\,\,\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{x}\end{array}$
$\Rightarrow {{\text{m}}_{\text{KL}\,\text{trc}}}\text{-}{{\text{m}}_{\text{KL}\,\text{sau}}}\text{=}{{\text{m}}_{\text{Fe}\,\text{pu}}}\text{-}{{\text{m}}_{\text{Cu}\,\text{tao}\,\text{ra}}}\Rightarrow \text{20-12,4=56}\text{.(0,15+x)-64x}$
$\Rightarrow \text{x=0,1}\,\text{mol}\,\Rightarrow \,{{\text{n}}_{\text{Cu(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\,\text{bd}}}\text{=0,4}\,\text{mol}\,\Rightarrow \,\text{m=86,9}\,\text{gam}$
