\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{150.36,5\%}{36,5}=1,5\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,4}{2}< \dfrac{1,5}{6}\), ta được HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{3}{2}n_{Al}=0,6\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,4\left(mol\right)\\n_{HCl\left(dư\right)}=3n_{Al}=1,2\left(mol\right)\end{matrix}\right.\)
⇒ VH2 = 0,6.24,79 = 14,874 (l)
nHCl (dư) = 1,5 - 1,2 = 0,3 (mol)
Ta có: m dd sau pư = 10,8 + 150 - 0,6.2 = 159,6 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,4.133,5}{159,6}.100\%\approx33,5\%\\C\%_{HCl}=\dfrac{0,3.36,5}{159,6}.100\%\approx6,86\%\end{matrix}\right.\)