PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a) Ta có: \(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)=n_{AlCl_3}=\frac{2}{3}n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)
b) Theo PTHH: \(n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow C_{M_{HCl}}=\frac{0,6}{0,2}=3\left(M\right)\)