Bài giải:
Ta có: $m_{kim..loại}+m_{Cl^-}=m_{muối}$
`⇔13,2+m_{Cl^-}=28,84`
`⇔m_{Cl^-}=15,64(g)`
`⇒n_{Cl^-}=\frac{15,64}{35,5}≈0,44(mol)`
`⇒n_{HCl}=n_{Cl^-}=0,44(mol)`
`-m_{chất..tan..HCl}=0,44.36,5=16,06(g)`
`⇒m_{dd..HCl}=\frac{16,06.100%}{10%}=160,6(g)`