\(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03(mol)\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ \Rightarrow n_{FeCl_3}=0,06(mol);n_{HCl}=0,18(mol)\\ \Rightarrow m_{FeCl_3}=0,06.162,5=9,75(g)\\ m_{dd_{HCl}}=\dfrac{0,18.36,5}{10,95\%}=60(g)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{60}{1,25}=48(ml)\)