\(m_{HCl}=13,035.0,1.1,05=1,36875\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{1,36875}{36,5}=0,0375\left(mol\right)\)
\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)
0,0375/2y---0,0375
Ta có: \(M_{Fe_xO_y}=\dfrac{1}{\dfrac{0,0375}{2y}}=\dfrac{160y}{3}\)
\(\Leftrightarrow56x+16y=\dfrac{160y}{3}\)
\(\Rightarrow x=2;y=3\)