Gọi cthc: FexOy ; x,y \(\in Z^+\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_{HCl}=\dfrac{16,425}{36,5}=0,45\left(mol\right)\)
Pt:\(Fe_xO_y+yCO\underrightarrow{t^o}xFe+yCO_2\uparrow\) (1)
\(\dfrac{0,15}{x}\)<---------------- 0,15
\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\) (2)
\(\dfrac{0,45}{2y}\)<---- 0,45
(1)(2) \(\Rightarrow\dfrac{0,15}{x}=\dfrac{0,45}{2y}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\)
Vậy cthc: Fe2O3