Điện phân dung dịch X chứa x mol

Chọn đáp án C

Giải thích:

$\begin{array}{l}+Thutukhutrencatot:C{u}^{2+}>H_{2}O;Thutuoxihoa:C{l}^{-}>H_{2}O.\\ +Fe+ddX\stackrel{}{\to }NO\uparrow \Rightarrow Xgom\left\{\begin{array}{l}N{O_3}^{-}:2xmol\\ N{a}^{+}:\mathrm{0,2}mol\\ H^{+}\\ C{u}^{2+}\left(cotheconhoachet\right)\end{array}\right\}vaC{l}^{-}\text{\hspace{0.17em}\hspace{0.17em}da\hspace{0.17em}\hspace{0.17em}}het.\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}{n}_{O_2}=a\\ n_{C{l}_2}=\mathrm{0,5}{n}_{C{l}^{-}}=\mathrm{0,1}\\ BTE:2n_{Cu}=2n_{C{l}_2}+4n_{O_2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Cu}=\mathrm{0,1}+2a\\ m_{ddgiam}=64(\mathrm{0,1}+2a)+32a+\mathrm{0,1.71}=\mathrm{21,5}\end{array}\right.\\ \Rightarrow a=\mathrm{0,05}\Rightarrow ddsaudienphanco:n_{H^{+}}=4n_{O_2}=\mathrm{0,2};n_{C{u}^{2+}}=x-\mathrm{0,2.}\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}N{O_3}^{-}:2x\\ N{a}^{+}:\mathrm{0,2}\\ H^{+}:\mathrm{0,2}\\ C{u}^{2+}:x-\mathrm{0,2}\end{array}\right\}+Fe\stackrel{}{\to }\underset{\frac{n_{H^{+}}}{4}=\mathrm{0,05}}{\underbrace{NO\uparrow }}+\left\{\begin{array}{l}N{O_3}^{-}:2x-\mathrm{0,05}\\ N{a}^{+}:\mathrm{0,2}\\ BTĐT\Rightarrow F{e}^{2+}:x-\mathrm{0,125}\end{array}\right\}+\underset{x-\mathrm{0,2}}{\underbrace{Cu\downarrow }}\\ \Rightarrow \Delta m=m_{Fepö}-m_{Cutaothanh}=56.(x-\mathrm{0,125})-64.(x-\mathrm{0,2})=\mathrm{1,8}\Rightarrow x=\mathrm{0,5.}\\ +n_{electrontraodoi}=\frac{11580.10}{96500}=\mathrm{1,2}mol;\left\{\begin{array}{l}BTE\text{\hspace{0.17em}\hspace{0.17em}o\hspace{0.17em}\hspace{0.17em}}catot:2\underset{\mathrm{0,5}}{\underbrace{n_{Cu}}}+2n_{H_2}=\mathrm{1,2}\\ BTEoanot:2\underset{\mathrm{0,1}}{\underbrace{n_{C{l}_2}}}+4n_{O_2}=\mathrm{1,2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_2}=\mathrm{0,1}\\ n_{O_2}=\mathrm{0,25}\end{array}\right.\\ \Rightarrow n_{khí}=n_{C{l}_2}+n_{H_2}+n_{O_2}=\mathrm{0,45}mol\Rightarrow V=\mathrm{10,08}lít.\end{array}$