FexOy+2yHCl\(\rightarrow\)\(FeCl_{\dfrac{2y}{x}}\)+yH2O
số mol HCl=0,4mol
số mol oxit=\(\dfrac{0,4}{2y}=\dfrac{0,2}{y}mol\)
(56x+16y)\(\dfrac{0,2}{y}=14,4\)
\(\dfrac{56x}{y}\)+16=72 suy ra \(\dfrac{56x}{y}\)=56 suy ra\(\dfrac{x}{y}=\dfrac{56}{56}=1\)
CTHH oxit FeO