2X+2xHCl--->2XClx+xH2
n X=\(\frac{11,2}{X}\left(mol\right)\)
n\(_{XCL_x}=\frac{25,4}{X+35,5x}\left(mol\right)\)
Theo pthh
n X=n XClx
--> \(\frac{11,2}{X}=\frac{25,4}{X+35,5x}\Leftrightarrow25,4X=11,2X+397,6x\)
\(\Leftrightarrow14,2X=397,6x\Leftrightarrow X=28x\)
x=2--->X=56(Fe)
Vậy X là Fe