Obsah železa v magnetitu
Kolik procent železa obsahuje ruda:
magnetite \( \mathrm{Fe}_{3} \mathrm{O}_{4} \)
\( \mathrm{f}\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)=\frac{3 \cdot M_{r}(F e)}{M_{r}\left(F e_{3} O_{4}\right)}=\frac{167,55}{231,55}=0,7236 \Rightarrow \mathbf{7 2 , 3 6} \% \mathbf{F e} \)
\( \mathrm{f}\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)=\frac{3 \cdot M_{r}(F e)}{M_{r}\left(F e_{3} O_{4}\right)}=\frac{167,55}{231,55}=0,7236 \Rightarrow \mathbf{8 0 , 0 0} \% \mathbf{F e} \)
\( \mathrm{f}\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)=\frac{3 \cdot M_{r}(F e)}{M_{r}\left(F e_{3} O_{4}\right)}=\frac{167,55}{231,55}=0,7236 \Rightarrow \mathbf{7 0 , 0 0} \% \mathbf{F e} \)
\( \mathrm{f}\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)=\frac{3 \cdot M_{r}(F e)}{M_{r}\left(F e_{3} O_{4}\right)}=\frac{167,55}{231,55}=0,7236 \Rightarrow \mathbf{6 5 , 5 5} \% \mathbf{F e} \)