PTHH: \(RO+H_2\rightarrow R+H_2O\\ 0,06mol:0,06mol\rightarrow0,06mol:0,06mol\)
\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
\(M_{RO}=\dfrac{13,38}{0,06}=223\left(g/mol\right)\)
\(M_R+16=223\Leftrightarrow m_R=207\left(g/mol\right)\)
Vậy R là Chì, KHHH Pb.