Đáp án:

 D

Giải thích các bước giải:

Hỗn hợp khí gồm \[C{O_2};{H_2}\] , \[{\overline M _{hhkhi}} = 35,6\]

Phương pháp đường chéo: \[\begin{array}{*{20}{c}} {44}&{}&{}&{}&{33,6} \\ {}& \searrow &{}& \nearrow &{} \\ {}&{}&{35,6}&{}&{} \\ {}& \nearrow &{}& \searrow &{} \\ 2&{}&{}&{}&{8,4} \end{array}\]

\[ \Rightarrow \dfrac{{{n_{C{O_2}}}}}{{{n_{{H_2}}}}} = \dfrac{{33,6}}{{8,4}} = \dfrac{4}{1} \Rightarrow \left\{ \begin{gathered}
  {n_{C{O_2}}} = 0,16 \hfill \\
  {n_{{H_2}}} = 0,04 \hfill \\ 
\end{gathered}  \right.\]

Dung dịch Y gồm: \[N{a^ + }(x{\text{ mol)}};{K^ + }{\text{ (y mol)}};C{l^ - }{\text{ (z mol)}}\]

Ta có: \[x + y = z\]

\[{m_{muoi}} = {m_{N{a^ + }}} + {m_{{K^ + }}} + {m_{C{l^ - }}} = 23x + 39y + 35,5\left( {x + y} \right) = 58,5x + 74,5y=52,46\] (1)

\[{m_{N{a_2}O}} = \dfrac{x}{2}.62 = 31x \Rightarrow {m_X} = \dfrac{{31x}}{{0,20741}} = 149,5x\]

\[{n_{HCl}} = {n_{C{l^ - }}} = x + y \Rightarrow {m_{{\text{dd}}HCl}} = \dfrac{{36,5(x + y)}}{{0,146}} = 250\left( {x + y} \right)\]

Bảo toàn khối lượng:

\[\begin{gathered}
  {m_X} + {m_{ddHCl}} = {m_{ddY}} + {m_{hhkhi}} \hfill \\
   \Leftrightarrow 149,5x + 250\left( {x + y} \right) = {m_{ddY}} + 0,2.35,6 \hfill \\
   \Rightarrow {m_{ddY}} = 399,5x + 250y - 7,12 \hfill \\ 
\end{gathered} \]

\[C{\% _{KCl}} = \dfrac{{{m_{KCl}}}}{{{m_{ddY}}}}.100\% = \dfrac{{74,5y}}{{399,5x + 250y - 7,12}}.100\% = 16,799\% \;{\text{(2)}}\]

Từ (1) và (2) ⇒ \[\left\{ \begin{gathered}
  x = 0,26 \hfill \\
  y = 0,5 \hfill \\ 
\end{gathered}  \right.\]

\[\begin{gathered} \Rightarrow {m_X} = 38,87;{n_K} = 2{n_{{H_2}}} = 0,08{\text{ (mol)}} \hfill \\ \Rightarrow \% {m_K} = \dfrac{{0,08.39}}{{38,87}}.100\% \approx 8,03\% \hfill \\ \end{gathered} \]