\(CTPT:R_2O_x\\ PTHH:R_2O_x+2xHCl\rightarrow2RCl_x+xH_2O\\ n_{oxit}=\frac{8}{2R+16x}\left(mol\right)\\ n_{HCl}=\frac{10,98}{36,5}=0,3\left(mol\right)\\ \rightarrow n_{oxit}=\frac{0,15}{x}\\ Theo\cdot pt:\frac{4}{R+8x}=\frac{0,15}{x}\\ \Leftrightarrow R=18,6x\\ BL:x=3\rightarrow R=56\left(Fe\right)\\ \rightarrow CTHH:Fe_2O_3\)
\(PTHH:A+2HCl\rightarrow ACl_2+H_2\)
\(n_{HCl}=0,4\left(mol\right);n_A=\frac{13}{A}\left(mol\right)\)
Theo đề ta có: \(\frac{13}{A}=\frac{0,4}{2}\Leftrightarrow A=65\left(Zn\right)\)