`n_(O_2)=\frac{4,8}{36}=0,15(mol)`
`n_(H_2)=\frac{3,36}{22,4}=0,15(mol)`
$4M+nO_2\xrightarrow{t^o}2M_2O_n$
`\frac{0,6}{n}` `0,15` `\frac{0,3}{n}`
`M_2O_n+2nHCl->2MCl_n+nH_2O`
`\frac{0,3}{n}` `\frac{0,6}{n}`
`2M+2nHCl->2MCl_n+nH_2`
`\frac{0,3}{n}` `\frac{0,3}{n}` `0,15`
`=>n_(M)=\frac{0,9}{n}`
`=>M_(M)=\frac{8,1}{\frac{0,9}{n}}=9n`
`=>n=1=>M=9(loại)`
`n=2=>M=18(loại)`
`n=3=>M=27(Al)`
Vậy `M` là `Al`
`b,`
`n_(Al(OH)_3)=\frac{15,6}{78}=0,2(mol)`
Theo `PT`
`n_(AlCl_3)=0,3(mol)`
`n_(HCl (pư))=n.n_(AlCl_3)=3.0,3=0,9(mol)`
`=>n_(HCl (dư))=0,9.110%-0,9=0,09(mol)``
Ta có
`n_(Al(OH)_3)<n_(AlCl_3)`
`=>` Có 2 trường hợp
`TH1:` `AlCl_3` dư
`HCl+NaOH->NaCl+H_2O`
`0,09` `0,09`
`AlCl_3+3NaOH->Al(OH)_3+3NaCl`
`0,2` `0,6` `0,2`
`=>V_(NaOH)=\frac{0,69}{2}=0,345(l)`
`TH2:` Kết tủa tan 1 phần
`HCl+NaOH->NaCl+H_2O`
`0,09` `0,09`
`AlCl_3+3NaOH->Al(OH)_3+3NaCl`
`0,3` `0,9` `0,3`
`Al(OH)_3+NaOH->NaAlO_2+2H_2O`
`0,1` `0,1`
`=>V_(NaOH)=\frac{1,09}{2}=0,545(l)`