Gọi: nAl = 2a (mol), nA = 3a (mol)
Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(A+2HCl\rightarrow ACl_2+H_2\)
Theo PT: nHCl = 2nH2 = 1,2 (mol)
\(\Rightarrow m_{ddHCl}=\dfrac{1,2.36,5}{21,9\%}=200\left(g\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_A\) ⇒ 0,6 = 3/2.2a + 3a ⇒ a = 0,1
⇒ nAl = 0,2 (mol), nA = 0,3 (mol)
⇒ 0,2.27 + 0,3.MA = 12,6 ⇒ MA = 24 (g/mol)
→ A là Mg.
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\end{matrix}\right.\)
Ta có: m dd sau pư = 12,6 + 200 - 0,6.2 = 211,4 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{211,4}.100\%\approx12,6\%\\C\%_{MgCl_2}=\dfrac{0,3.95}{211,4}.100\%\approx13,5\%\end{matrix}\right.\)
