\(m_{Fe_3O_4}=1.81,2\%=0,812\left(tấn\right)\)
\(Có:n_{Fe_3O_4}=\dfrac{0,812.1000000}{232}=3500\left(mol\right)\)
\(\Rightarrow n_{Fe_{\left(Fe_3O_4\right)}}=3n_{Fe_3O_4}=3.3500=10500\left(mol\right)\)
\(\Rightarrow m_{Fe}=10500.56=588000\left(g\right)=588\left(kg\right)\)