\(Fe+2HCL\rightarrow FeCl_2+H_2\left(1\right)\)
0,3___0,6____________0,3
\(Fe_xO_y+2yHCl\rightarrow xFeCl2y/x+yH_2O\left(2\right)\)
0,4/y_______ 0,8_________________________
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\%m_{Fe}=\frac{0,3.56}{40}.100\%=42\%,\%m_{FexOy}=100-42=58\%\%\)
\(m_{HCl}=400.16,425=65,7\left(g\right)\)
\(\rightarrow n_{HCl}=1,8\left(mol\right)\)
\(C\%_{HCl_{du}}=\frac{m}{m_{dd}}\Leftrightarrow2,92\%=\frac{m_{HCl_{du}}.36,5}{40+400+60,6+0,3.2}\)
\(\Leftrightarrow14,6=n_{HCl_{du}}.36,5\)
\(\Leftrightarrow n_{HCl_{du}}=0,4\left(mol\right)\)
\(n_{HCL\left(2\right)}=1,8-0,6-0,4=0,8\left(mol\right)\)
\(m_{FexOy}=40-0,3.56=23,2\left(g\right)\)
\(56x+16y=\frac{23,2}{\frac{0,4}{y}}\Leftrightarrow56x+16y=58y\)
\(\Leftrightarrow56x=42y\Leftrightarrow x:y=3:4\)
Vậy CTHH là Fe3O4