$m_{Fe}=1.98\%=0,98$ tấn
Trong $Fe_2O_3.2H_2O$, $\%Fe=\dfrac{56.2.100}{160+18.2}=\dfrac{400}{7}\%$
$\Rightarrow m_{Fe_2O_3.2H_2O}=0,98:\dfrac{400}{7}\%=1,715$ tấn
$\Rightarrow m_{\text{quặng (lí thuyết)}}=1,715:80\%=2,14375$ tấn
$\to H=\dfrac{2,14375.100}{2,305}=93\%$