Hỗn hợp $Y : \begin{cases}MgO: \ x(mol)\\Al_2O_3: \ y(mol)\\CuO: \ z(mol)\\\end{cases}$

$2Mg+O_2\xrightarrow{t^o}2MgO$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$2Cu+O_2\xrightarrow{t^o}2CuO$

`MgO+2HCl->MgCl_2+H_2`

`Al_2O_3+6HCl->2AlCl_3+3H_2O`

`CuO+2HCl->CuCl_2+H_2O`

Ta thấy

`O_2->2H_2O`

`->n_{O_2}=1/2 n_{H_2O}`

`2HCl->H_2O`

`=>O_2->4HCl`

Theo phương trình

`->n_{O_2}=4n_{Cl}`

Lại có

`m_{\text{oxit}}=16,2(g)`

`=>m+m_{O_2}=16,2(g)`

`->m+32x=16,2(1)`

`m_{\text{Muối}}=38,2(g)`

`=>m+m_{Cl}=38,2`

`=>m+142x=38,2(2)`

`(2)-(1)`

`->110x=22(g)`

`->x=0,2(mol)`

`->m_{O_2}=0,2.32=6,4(g)`

`->m=16,2-6,4=9,8(g)`

`b,`

Ta có

`16,2(g) \ Y` tác dụng hết với `0,8(mol) \ H^{+}`

Gọi `V_{\text{ dd axit}}=V`

`->1V+2.0,5V=0,8`

`->2V=0,8(mol)`

`->V=0,4(mol)`

`->m_{Cl^-}+m_{SO_4^{2-}}=0,4.35,5+0,4.96=52,6(g)`

`->a=m+m_{Cl^-}+m_{SO_4^{2-}}=9,8+52,6=62,4(g)`

Cho `n_{HCl}=a(mol)`

`=> n_{H_2O}=0,5a(mol)`

`a)`BTKL:

`m_{Y}+m_{HCl}=m_{\text{muối}}+m_{H_2O}`

`=> 16,2+36,5a=38,2+9a`

`=> a=0,8(mol)`

`=> n_{Cl^{-}}=0,8(mol)`

`m_{\text{muối}}=m_{\text{kl}}+m_{Cl^{-}}`

`=>38,2=m + 0,8.35,5`

`=> m=9,8g`

`b)` Do chỉ tạo ra dung dịch chứa muối 1`=>` axit phản ứng hết

Mặc khác do phần `a` , `16,2g` `Y` phản ứng hết với `0,8(mol)` axit

`=> n_{H^{+}}=0,8(mol)`

`=> 1V+1,5V=0,8`

`=> V=0,4(l)`

`=> n_{HCl}=0,4(mol)`

`n_{H_2SO_4}=0,4(mol)`

`m_{\text{muối}}=m_{\text{kl}}+m_{SO_4^{2-}}+m_{Cl^{-}}`

`=> m_{\text{muối}}=9,8+0,4.(96+35,5)=62,4g`