Tiến hành điện phân dung dịch chứa 0,25 mol Cu(NO3)­2 và 0,18 mol NaCl bằng điện cực trơ, màng ngăn xốp với cường �

\(\begin{array}{l}

C{u^{2 + }} + 2e \to Cu\\

a{\rm{   }} \to 2a \to a

\end{array}\)                                                  

\(\begin{array}{l}

2C{l^ - } \to C{l_2} + 2e\\

0,18 \to 0,09 \to 0,18\\

2{H_2}O \to 4{H^ + } + {O_2} + 4e\\

{\rm{                          }}b{\rm{     }}4b

\end{array}\)

\(\left\{ \begin{array}{l}

2a = 0,18 + 4b\\

64a + 0,09.71 + 32b = 21,75

\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}

a = 0,21\\

b = 0,06

\end{array} \right.\)

\(\left\{ \begin{array}{l}

C{u^{2 + }}\left( {0,04} \right)\\

{H^ + }\left( {0,24} \right)\\

NO_3^ - \left( {0,5} \right)\\

N{a^ + }\left( {0,18} \right)

\end{array} \right.\left\{ \begin{array}{l}

CR\left\{ \begin{array}{l}

Fe\left( z \right)\\

Cu\left( {0,04} \right)

\end{array} \right.\\

{\rm{dd}}\left\{ \begin{array}{l}

N{a^ + }\\

F{e^{2 + }}\left( x \right)\\

NO_3^ -

\end{array} \right.\\

NO\left( y \right)\\

{H_2}O

\end{array} \right.\)

BTBT(Cu): \({n_{Cu}} = 0,04\)

BT e: 2x = 0,04.2 + 3y

BTNT (N): \({n_{NO_3^ - \left( {{\rm{dd}}} \right)}} = 0,5 – y\)

BTDT: 2x + 0,18 = 0,5 - y

\( \to \left\{ \begin{array}{l}

x = 0,13\\

y = 0,06

\end{array} \right.\)

\(56z + 0,04.64 = 0,75\left( {56x + 56z} \right)\)

\( \to z = \frac{{29}}{{140}} \to m = \left( {0,13 + \frac{{29}}{{140}}} \right)56 = 18,88\left( g \right)\)