a: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b: \(n_{AlCl_3}=\dfrac{26.7}{27+35.5\cdot3}=0.2\left(mol\right)\)
=>nAl=0,2(mol)
\(m=0.2\cdot27=5.4\left(g\right)\)
c: \(2\cdot n_{Al}=3\cdot n_{H_2}\Leftrightarrow n_{H_2}=\dfrac{2}{3}\cdot\dfrac{1}{5}=\dfrac{2}{15}\left(mol\right)\)
\(V=\dfrac{2}{15}\cdot22.4=\dfrac{224}{75}\left(lít\right)\)