\(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right);n_{Cl_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ PTHH:H_2+Cl_2\rightarrow\left(as\right)2HCl\\ Vì:\dfrac{1}{1}>\dfrac{1,5}{1}\Rightarrow Cl_2dư\\ n_{HCl\left(thu.được\right)}=75\%.n_{H_2}=75\%.1=0,75\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{Fe_2O_3}=\dfrac{n_{HCl}}{6}=\dfrac{0,75}{6}=0,125\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=160.0,125=20\left(g\right)\)


