Gọi công thức muối clorua đó là: \(RCl_n\)
\(m_{ct}=60.5,35\%=3,21\)
\(\Rightarrow n_{RCl_n}=\dfrac{3,21}{R+35,5n}\)
\(RCl_n\left(\dfrac{3,21}{R+35,5n}\right)+nAgNO_3\rightarrow R\left(NO_3\right)_n+nAgCl\left(\dfrac{3,21n}{R+35,5n}\right)\)
\(\Rightarrow\dfrac{3,21n}{R+35,5n}=2.0,03\)
\(\Leftrightarrow R=18n\)
Thế \(n=1;2;3;...\) ta nhận \(\left\{{}\begin{matrix}n=3\\R=54\end{matrix}\right.\)