`Mg+2HCl->MgCl_2+H_2`
`2Al+6HCl->2AlCl_3+3H_2`
`n_(H_2)=\frac{8,96}{22,4}=0,4(mol)`
`=>m_(H_2)=0,4.2=0,8(mol)`
Theo `PT`
`n_(HCl)=2n_(H_2)=0,8(mol)`
`=>m_(HCl)=0,8.36,5=29,2(g)`
Áp dụng `ĐLBTKL`
`m_(hh)+m_(HCl)=m_(Muối)+m_(H_2)`
`=>m_(Muối)=7,8+29,2-0,8=36,2(g)`