\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ m_{Fe_3O_4}=11,4-0,1.56=5,8\left(g\right)\\ n_{Fe_3O_4}=\dfrac{5,8}{232}=0,025\left(mol\right)\\ Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\\ n_{HCl\left(tổng\right)}=2.n_{Fe}+8.n_{Fe_3O_4}=2.0,1+8.0,025=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{1,25}=0,32\left(l\right)\)