Đáp án:

\({V_{{H_2}}} = 8,96l\)

Giải thích các bước giải:

Bảo toàn khối lượng ta có:

\(\begin{array}{l}
2Mg + {O_2} \to 2MgO\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
Mg + C{l_2} \to MgC{l_2}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
{m_X} + {m_Y} = {m_{hỗn hợp rắn}}
\end{array}\)

\(\begin{array}{l}
 \to {m_{{O_2}}} + {m_{C{l_2}}} = 25,2 - 7,8 = 17,4g\\
 \to 32{n_{{O_2}}} + 71{n_{C{l_2}}} = 17,4
\end{array}\)

Mặt khác ta có: \({n_{{O_2}}} + {n_{C{l_2}}} = 0,3mol\)

Giải hệ phương trình sau:

\(\begin{array}{l}
\left\{ \begin{array}{l}
32{n_{{O_2}}} + 71{n_{C{l_2}}} = 17,4\\
{n_{{O_2}}} + {n_{C{l_2}}} = 0,3
\end{array} \right.\\
 \to {n_{{O_2}}} = 0,1mol \to {n_{C{l_2}}} = 0,2mol
\end{array}\)

Bảo toàn electron ta có:

\(\begin{array}{l}
Mg \to M{g^{2 + }} + 2e\\
A{l^{3 + }} \to A{l^{3 + }} + 3e
\end{array}\)

\(\begin{array}{l}
{O_2} + 4e \to 2{O^{2 - }}\\
C{l_2} + 2e \to 2C{l^ - }
\end{array}\)

\(\begin{array}{l}
 \to 2{n_{Mg}} + 3{n_{Al}} = 4{n_{{O_2}}} + 2{n_{C{l_2}}}\\
 \to 2{n_{Mg}} + 3{n_{Al}} = 0,8mol
\end{array}\)

Ta lại có:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg \to M{g^{2 + }} + 2e\\
A{l^{3 + }} \to A{l^{3 + }} + 3e\\
2{H^ + } + 2e \to {H_2}
\end{array}\)

\(\begin{array}{l}
2{n_{Mg}} + 3{n_{Al}} = 2{n_{{H_2}}}\\
 \to {n_{{H_2}}} = 0,4mol\\
 \to {V_{{H_2}}} = 8,96l
\end{array}\)