Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(m_{HCl}=438.10\%=43,8\left(g\right)\Rightarrow n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{1,2}{6}\), ta được HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=1,2-0,6=0,6\left(mol\right)\)
Ta có: m dd sau pư = 5,4 + 438 - 0,3.2 = 442,8 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{442,8}.100\%\approx6,03\%\\C\%_{HCl}=\dfrac{0,6.36,5}{442,8}.100\%\approx4,95\%\end{matrix}\right.\)
