Bài 5:
a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
b) Ta có: \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\) \(\Rightarrow n_{Al}=0,04mol\)
\(\Rightarrow m_{Al}=0,04\cdot27=1,08\left(g\right)\) \(\Rightarrow\%m_{Al}=\dfrac{1,08}{3,12}\cdot100\%\approx34,62\%\)
\(\Rightarrow\%m_{Al_2O_3}=65,38\%\)