Gọi CTHH oxit sắt là $Fe_xO_y$
- P1:
$n_{HCl}=0,15.1,5=0,225(mol)$
$Fe_xO_y+2yHCl\to xFeCl_{\frac{2y}{x}}+yH_2O$
$\to n_{Fe_xO_y}=\dfrac{0,225}{2y}=\dfrac{0,1125}{y}(mol)$
- P2:
$n_{Fe}=\dfrac{4,2}{56}=0,075(mol)$
$Fe_xO_y+yH_2\xrightarrow{{t^o}} xFe+yH_2O$
$\to n_{Fe_xO_y}=\dfrac{0,075}{x}(mol)$
Do đó ta có:
$\dfrac{0,075}{x}=\dfrac{0,1125}{y}$
$\to \dfrac{x}{y}=\dfrac{0,075}{0,1125}=\dfrac{2}{3}$
Vậy CTHH oxit sắt là $Fe_2O_3$