Điện phân điện cực trơ màng ngăn xốp, dung dịch X chứa a gam Cu(NO3)2 và b gam NaCl

* Hướng dẫn giải

Đáp án D

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \underset{Z, anot}{\underbrace{\left\{\begin{array}{l}C{l}_2\uparrow \\ O_2\uparrow \end{array}\right\}}}: 0,51 mol\\ \underset{X}{\underbrace{\left\{\begin{array}{l}Cu(N{O}_3{)}_2: x mol\\ NaCl: y mol\end{array}\right\}}} \underset{Y}{\underbrace{\left\{\begin{array}{l}N{a}^{+}, H^{+}\\ N{O_3}^{-}\end{array}\right\}}}\underset{T}{\underbrace{\underset{max}{\overset{Fe}{\to }}\left\{\begin{array}{l}F{e}^{2+}: 0,225\\ N{a}^{+}, N{O_3}^{-}\end{array}\right\}}}+NO\uparrow \\ \\ \underset{catot}{\underbrace{Cu}}\\ + Y+Fe: \left\{\begin{array}{l}BTE: 2n_{Fe}=3n_{NO}\\ n_{H^{+}}=4n_{NO}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=0,15\\ n_{H^{+}}=0,6\end{array}\right.\Rightarrow n_{H_{2}O bị oxi hóa}=\frac{n_{H^{+}}}{2}=0,3\\ \Rightarrow Z còu \left\{\begin{array}{l}{n}_{O_2}=\frac{n_{H_{2}O bị oxi hóa }}{2}=0,15\\ n_{C{l}_2}=0,51-0,15=0,36\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{N{a}^{+}}=n_{C{l}^{-}}=2n_{C{l}_2}=0,72\\ BTĐT: n_{N{O_3}^{-} trong Y}=0,72+0,6=1,32\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}{n}_{NaCl}=0,72\\ n_{Cu{\left(N{O}_3\right)}_2}=0,66\end{array}\right.\Rightarrow m_{hỗn hợp}=166,2\end{array}$