\(n_{HCl}=0,15.3=0,45\left(mol\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
Gọi cthc: FexOy , x,y \(\in Z^+\)
\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)(1)
0,45/2y <---- 0,45
\(Fe_xO_y+yCO\rightarrow xFe+yCO_2\) (2)
0,15/x <--------------0,15
(1)(2) \(\dfrac{0,45}{2y}=\dfrac{0,15}{x}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\)
Vậy cthc: Fe2O3