Obsah železa v rudě krevel
Kolik procent železa obsahuje ruda:
krevel, \( \mathrm{Fe}_{2} \mathrm{O}_{3} \)
\( \mathrm{f}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=\frac{2 \cdot M_{r}(\mathrm{Fe})}{M_{r}\left(Fe_{2} \mathrm{O}_{3}\right)}=\frac{111,7}{159,7}=0,6994 \Rightarrow \mathbf{59,94} \% \mathbf{Fe} \)
\( \mathrm{f}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=\frac{2 \cdot M_{r}(\mathrm{Fe})}{M_{r}\left(Fe_{2} \mathrm{O}_{3}\right)}=\frac{111,7}{159,7}=0,6994 \Rightarrow \mathbf{69,94} \% \mathbf{Fe} \)
\( \mathrm{f}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=\frac{2 \cdot M_{r}(\mathrm{Fe})}{M_{r}\left(Fe_{2} \mathrm{O}_{3}\right)}=\frac{111,7}{159,7}=0,6994 \Rightarrow \mathbf{49,94} \% \mathbf{Fe} \)
\( \mathrm{f}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=\frac{2 \cdot M_{r}(\mathrm{Fe})}{M_{r}\left(Fe_{2} \mathrm{O}_{3}\right)}=\frac{111,7}{159,7}=0,6994 \Rightarrow \mathbf{79,94} \% \mathbf{Fe} \)