\(n_{Al}=\dfrac{m}{M}=\dfrac{4,05}{27}=0,15\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
Theo PTHH: \(n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\)
\(V_{H_2\left(\text{đ}ktc\right)}=n.22,4=0,225.22,4=5,04\left(l\right)=5040\left(ml\right)\)
Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\)
\(m_{AlCl_3}=n.M=0,15.133,5=20,025\left(g\right)\)