\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \Rightarrow n_{Fe}=0,1\left(mol\right)\\ \Rightarrow m_{Fe}=0,1\cdot56=5,6\left(g\right)\\ \Rightarrow m_{Fe_2O_3}=m_{hh}-m_{Fe}=16\left(g\right)\\ \Rightarrow n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ \Rightarrow\sum n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=0,2+0,6=0,8\left(mol\right)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,4}=2M\)
Chọn A