FexOy + 2HCl \(\rightarrow\) xFe2y/x + yH2O
nFexOy = \(\dfrac{16}{56x+16y}\) ( mol ) (1)
Theo PTHH ta có: nFe = \(\dfrac{0,3}{y}\) ( mol ) (2)
Từ (1)(2) ta có
\(\dfrac{16}{56x+16y}=\dfrac{0,3}{y}\)
\(\Leftrightarrow\) \(16y=16,8x+4,8y\)
\(\Leftrightarrow\) \(11,2y=16,8x\)
\(\Leftrightarrow\) \(\dfrac{x}{y}=\dfrac{11,2}{16,8}=\dfrac{2}{3}\)
\(\Rightarrow\) CTHH: Fe2O3