\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ LTL:\dfrac{0,2}{2}>\dfrac{0,4}{6}\\ \Rightarrow Aldư\\ n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{AlCl_3}=\dfrac{2}{15}.133,5=17,8\left(g\right)\\ n_{H_2}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(n_{HCl}=0.2\cdot2=0.4\left(mol\right)\)
=>HCl dư
\(n_{AlCl_3}=n_{Al}=0.2\left(mol\right)\)
\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)
\(n_{HCl}=3\cdot0.2=0.6\left(mol\right)\)
hay \(n_{H_2}=0.3\left(mol\right)\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(lít\right)\)
