Đáp án:

$\%V_{O_2}=46,15\%$

Giải thích các bước giải:

\({n_{Mg}} = 0,08,{n_{Fe}} = 0,08mol\)

\(\left\{ \begin{gathered}
  Mg \hfill \\
  Fe \hfill \\ 
\end{gathered}  \right.\xrightarrow{{ + {O_2},C{l_2}}}\left\{ \begin{gathered}
  MgO,MgC{l_2} \hfill \\
  FeO,F{e_2}{O_3},F{e_3}{O_4} \hfill \\
  FeC{l_2},FeC{l_3} \hfill \\ 
\end{gathered}  \right.\xrightarrow{{ + 0,24\ molHCl}}d{\text{d Z}}\left\{ \begin{gathered}
  MgC{l_2}:0,08{\text{ }}mol \hfill \\
  FeC{l_2}:a\ mol \hfill \\
  FeC{l_3}:b\ mol \hfill \\ 
\end{gathered}  \right.\)

Tiếp tục cho AgNO3 dư vào dung dịch Z:

\(A{g^ + } + C{l^ - } \to AgCl \downarrow \)

\(F{e^{2 + }} + A{g^ + } \to F{e^{3 + }} + Ag \downarrow \)

BTNT Fe ta có: $a+b=0,08$                             (1)

Bảo toàn Cl: $n_{AgCl}=2n_{MgCl_2}+2n_{FeCl_2}+3n_{FeCl_3}$

$→n_{AgCl}=0,08.2+2a+3b$

Ta có: $m↓=n_{AgCl}+m_{Ag}$

$→143,5.(0,16+2a+3b)+108a=56,69$                 (2)

Giải hệ (1), (2): $a=0,02;\ b=0,06$

BTNT Cl: $2n_{Cl_2}+n_{HCl}=n_{AgCl}$

$→2n_{Cl_2}+0,24=(0,16+2.0,02+3.0,06)→n_{Cl_2}=0,07\ mol$

BTNT H: $n_{H_2O}=\dfrac{1}{2}.n_{HCl}=0,12\ mol$

BTNT O: $2n_{O_2}=n_{H_2O} →n_{O_2}=0,06\ mol$

$\%V_{O_2}=\dfrac{0,06}{0,13}.100\%=46,15\%$