\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}.0,6=0,9\left(mol\right);n_{HCl}=\dfrac{6}{2}.0,6=1,8\left(mol\right)\\ a,V=V_{ddHCl}=\dfrac{1,8}{0,2}=9\left(l\right)=9000\left(ml\right)\\ b,V_{H_2\left(đktc\right)}=0,9.22,4=20,16\left(l\right)\)
`2Al+6HCl->2AlCl_3+3H_2`
0,6----1,8---------------------0,9 mol
`n_Al=(16,2)/27=0,6 mol`
`->V_(HCL)=(1,8)/(0,2)=9 ml`
`->V_(H_2)=0,9.22,4=20,16l`
`#YBTran:3`