\(PTHH:RCO_3\underrightarrow{t^o}RO+CO_2\)
\(n_{RCO_3}=\frac{17,4}{R+60}\left(mol\right);n_{RO}=\frac{12}{R+16}\left(mol\right)\)
Theo pt: \(n_M=n_{oxit}\\ \Leftrightarrow\frac{17,4}{R+60}=\frac{12}{R+16}\\ \Leftrightarrow17,4R+278,4=12R+720\\ \Leftrightarrow5,4R=441,6\\ \Leftrightarrow R=82\)
\(\rightarrow R:Pb\) (chì)