PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (2)
a) Ta có: \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow n_{Al}=0,1mol\)
\(\Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\) \(\Rightarrow m_{Al_2O_3}=10,2\left(g\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Al}=n_{AlCl_3\left(1\right)}=0,1mol\\n_{AlCl_3\left(2\right)}=2n_{Al_2O_3}=2\cdot\frac{10,2}{102}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3\left(1\right)}=0,1\cdot133,5=13,35\left(g\right)\\m_{AlCl_3\left(2\right)}=0,2\cdot133,5=26,7\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{muối}=40,05\left(g\right)\)
c) Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=3n_{Al}=0,3mol\\n_{HCl\left(2\right)}=6n_{Al_2O_3}=1,2mol\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=1,5mol\) \(\Rightarrow V_{ddHCl}=\frac{1,5}{1}=1,5\left(l\right)\)
