Đáp án:

\(m = 4,16g\)

Giải thích các bước giải:

\(\begin{array}{l}
{P_1}:\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,344}}{{22,4}} = 0,06\,mol\\
 \Rightarrow 1,5{n_{Al}} + {n_{Zn}} + {n_{Mg}} = 0,06\,mol(1)\\
{P_2}:\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
{n_{{O_2}}} = 0,75{n_{Al}} + 0,5{n_{Mg}} + 0,5{n_{Zn}}(2)\\
(1),(2) \Rightarrow {n_{{O_2}}} = 0,03\,mol\\
BTKL:{m_{hhX}} + {m_{{O_2}}} = {m_{cr}}\\
 \Rightarrow {m_{hhX}} = 3,04 - 0,03 \times 32 = 2,08g\\
 \Rightarrow m = 2,08 \times 2 = 4,16g
\end{array}\)