Tiến hành điện phân dung dịch chứa m gam hỗn hợp Cu(NO3)2 và NaCl...

$\begin{aligned}

& + \left\{ \begin{aligned}

& I=2A \\

& tg=3378s \\

\end{aligned} \right.\Leftrightarrow {{n}_{electron . trao.doi}}=\dfrac{3378.2}{96500}=0,07 mol; \left\{ \begin{aligned}

& Catot: {{n}_{Cu}}=0,025\Rightarrow {{n}_{{{H}_{2}}}}=\dfrac{0,07-0,025.2}{2}=0,01 mol \\

& Anot: \left\{ \begin{aligned}

& {{n}_{C{{l}_{2}}}}+{{n}_{{{O}_{2}}}}=0,035-0,01=0,025 \\

& 2{{n}_{C{{l}_{2}}}}+4{{n}_{{{O}_{2}}}}=0,07 \\

\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}

& {{n}_{C{{l}_{2}}}}=0,015 \\

& {{n}_{{{O}_{2}}}}=0,01 \\

\end{aligned} \right. \\

\end{aligned} \right. \\

& \Rightarrow tg= t (s) anot . giai .phong. C{{l}_{2}}, catot . giai . phong. Cu\Rightarrow a=b (*)\Rightarrow {{n}_{electron. trao. doi}}=2a=2b. \\

& + tg=2t (s)\Rightarrow {{n}_{electron . trao . doi}}=4a=4b\Rightarrow \left\{ \begin{aligned}

& catot: {{n}_{{{H}_{2}}}}=\dfrac{4b-2(b+0,025)}{2}=b-0,025 \\

& anot: {{n}_{{{O}_{2}}}}=\dfrac{4a-2(a+0,015)}{4}=0,5a-0,0075 \\

\end{aligned} \right. \\

& \Rightarrow {{n}_{kh\acute{i} \hat{o}\hat{u} 2t (s)}}=(b-0,025)+(0,5a-0,0075)+(a+0,015)=2,0625a (**) \\

& + \left\{ \begin{aligned}

& (*) \\

& (**) \\

\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}

& a=0,04 \\

& b=0,04 \\

\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}

& {{n}_{Cu{{(N{{O}_{3}})}_{2}}}}=0,065 \\

& {{n}_{NaCl}}=2(0,04+0,015)=0,11 \\

\end{aligned} \right.\Rightarrow m=18,655\Rightarrow gan . nhat . voi \\

\end{aligned}$