2Al + 6HCl → 2AlCl3 + 3H2 (1)
Mg + 2HCl → MgCl2 + H2 (2)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
a) Gọi số mol của Al và Mg lần lượt là x, y
Ta có: \(\left\{{}\begin{matrix}1,5x+y=0,6\\27x+24y=12,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
\(\Rightarrow\%m_{Al}=\dfrac{5,4}{12,6}\times100\%=42,86\%\)
\(m_{Mg}=0,3\times24=7,2\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{7,2}{12,6}\times100\%=57,14\%\)
b) Theo PT1: \(n_{HCl}=3n_{Al}=3\times0,2=0,6\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{Mg}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,6+0,6=1,2\left(mol\right)\)
\(\Rightarrow V_{ddHCl.1M}=\dfrac{1,2}{1}=1,2\left(l\right)\)
