`2Al + 6HCl -> 2AlCl_3 + 3H_2 \uparrow`
`2/15` `0,4` `2/15` `0,2` `(mol)`
`n_[Al]=[5,4]/27=0,2(mol)`
`n_[HCl]=[7,3.200]/[100.36,5]=0,4(mol)`
Có: `[0,2]/1 > [0,4]/6->Al` dư
`@V_[H_2]=0,2.22,4=4,48(l)`
`@m_[AlCl_3]=2/15.133,5=17,8(g)`
`@C%_[AlCl_3]=[17,8]/[5,4+200-0,2.2].100=8,68%`