\(Fe_3O_4+4CO\rightarrow\left(t^o\right)3Fe+4CO_2\)
1/2 2 1,5 2 ( mol )
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
1,625 4,875 3,25 ( mol )
\(n_{Fe}=\dfrac{266}{56}=4,75mol\)
\(n_{CaCO_3}=\dfrac{200}{100}=2mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
2 2 ( mol )
\(n_{Fe\left(tạo.ra.từ.H_2\right)}=4,75-1,5=3,25mol\)
\(\left\{{}\begin{matrix}V_{CO}=2.22,4=44,8l\\V_{H_2}=4,875.22,4=109,2l\end{matrix}\right.\)
\(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{1}{2}.232=116g\\m_{Fe_2O_3}=1,625.160=260g\end{matrix}\right.\)
