Bạn tham khảo cách giải câu b nha
$12,6g\,\left\{ \begin{gathered} Mg\,x\,mol \hfill \\ Al\,y\,mol \hfill \\ \end{gathered} \right.\,\, + \left\{ \begin{gathered} C{l_2} \hfill \\ {O_2} \hfill \\ \end{gathered} \right.\,\, \to hhZ$
$hhZ\left\{ \begin{gathered} Mg\,Al \hfill \\ Cl\,\,\,\,O \hfill \\ \end{gathered} \right.\, + \,0.8\,mol\,HCl \to muoi\, + \,\left\{ \begin{gathered} 0.2mol\,{H_2} \hfill \\ 0.2\,mol\,{H_2}O\,\,(Bao\,toan\,H) \hfill \\ \end{gathered} \right.$
Kết tủa là AgCl
${n_ \downarrow } = 172.2/143.5 = 1.2\,mol$
$Bt\,Cl:\,{n_{AgCl}} = {n_{HCl}} + 2{n_{C{l_2}}} \to {n_{C{l_2}}} = \frac{{1.2 - 0.8}}{2} = 0.2\,mol$ $Bt\,O:2{n_{{O_2}}} = {n_{{H_2}O}} \to {n_{{O_2}}} = 0.1\,mol$
$Bt\,e:\,2{n_{Mg}} + 3{n_{Al}} = 2{n_{C{l_2}}} + 4{n_{{O_2}}} + 2{n_{{H_2}}}\, \Rightarrow \,2x + 3y = 1.2$
$24x + 27y = 12.6\, \to \left\{ \begin{gathered} x = 0.3 \hfill \\ y = 0.2 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} \% Mg = 57.14\% \hfill \\ \% Al = 42.86\% \hfill \\ \end{gathered} \right.$
Các phương trình có thể xảy ra:
$\begin{gathered}
Mg + C{l_2} \to MgC{l_2} \hfill \\
Al + 1,5C{l_2} \to AlC{l_2} \hfill \\
Mg + 0,5{O_2} \to MgO \hfill \\
Al + 1,5{O_2} \to A{l_2}{O_3} \hfill \\
MgO + 2HCl \to MgC{l_2} + {H_2}O \hfill \\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O \hfill \\
Mg + 2HCl \to MgC{l_2} + {H_2} \hfill \\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2} \hfill \\
MgC{l_2} + 2AgN{O_3} \to Mg{(N{O_3})_2} + 2AgCl \hfill \\
AlC{l_3} + 3AgN{O_3} \to Al{(N{O_3})_3} + 3AgCl \hfill \\
\end{gathered} $
