Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{500.14,6\%}{36,5}=2\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{2}{6}\), ta được HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)
⇒ nHCl (dư) = 2 - 0,6 = 1,4 (mol)
Ta có: m dd sau pư = 5,4 + 500 - 0,3.2 = 504,8 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{1,4.36,5}{504,8}.100\%\approx10,12\%\\C\%_{AlCl_3}=\dfrac{0,2.133,5}{504,8}.100\%\approx5,29\%\end{matrix}\right.\)