\(n_{HCl}=0,015.2=0,03mol\)
\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)
\(n_{Fe_xO_y}=\dfrac{1}{2y}.n_{HCl}=\dfrac{0,03}{2y}mol\)
56x+16y=\(\dfrac{0,8}{\dfrac{0,03}{2y}}=\dfrac{160y}{3}\)
hay: 168x+48y=160y \(\rightarrow\)168x=112y
\(\dfrac{x}{y}=\dfrac{112}{168}=\dfrac{2}{3}\)
Fe2O3