PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)
Có: m dd sau pư = 5,4 + 200 - 0,3.2 = 204,8 (g)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,2.133,5}{204,8}.100\%\approx13,04\%\)