Đáp án:

\({V_{dd{\text{ HCl}}}} = 0,5{\text{ lít}}\)

Giải thích các bước giải:

 Sơ đồ phản ứng:

\(\left\{ \begin{gathered}
  Mg \hfill \\
  Al \hfill \\
  Zn \hfill \\ 
\end{gathered}  \right.\xrightarrow{{ + {O_2}}}\left\{ \begin{gathered}
  MgO \hfill \\
  A{l_2}{O_3} \hfill \\
  ZnO \hfill \\ 
\end{gathered}  \right.\xrightarrow{{ + HCl}}\left\{ \begin{gathered}
  MgC{l_2} \hfill \\
  AlC{l_3} \hfill \\
  ZnC{l_2} \hfill \\ 
\end{gathered}  \right. + {H_2}O\)

BTKL:

\({m_{hh{\text{ kl}}}} + {m_{{O_2}}} = {m_{oxit}}\)

\(21 + {m_{{O_2}}} = 29 \to {m_{{O_2}}} = 8{\text{ gam}}\)

\( \to {n_{{O_2}}} = \frac{8}{{16.2}} = 0,25{\text{ mol}}\)

Bảo toàn \(O\)

\({n_{{H_2}O}} = 2{n_{{O_2}}} = 0,25.2 = 0,5{\text{ mol}}\)

Bảo toàn \(H\)

\({n_{HCl}} = 2{n_{{H_2}O}} = 0,5.2 = 1{\text{ mol}}\)

\( \to {V_{dd{\text{ HCl}}}} = \frac{1}{2} = 0,5{\text{ lít}}\)