3.
- Đặt CTHH dạng: \(Fe_X^{III}Cl_Y^I\) .
Ta có: III.x=I.y
=>\(\dfrac{x}{y}\)=\(\dfrac{I}{III}\)=\(\dfrac{1}{3}\)=>\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy CTHH là FeCl3
PTK FeCl3=56+ 35,5.3=162,5 đvC
- Đặt CTHH dạng: \(Fe_x^{III}\left(SO_4\right)_y^{II}\)
Ta có: III.x=II.y
=>\(\dfrac{x}{y}\)=\(\dfrac{II}{III}\)=\(\dfrac{2}{3}\)=>\(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy CTHH là Fe2Cl3
PTK Fe2Cl3=56.2+ 35,5.3=218,5 đvC
- - Đặt CTHH dạng:\(Fe_x^{III}\left(NO_3\right)_y^I\)
Ta có: III.x=I.y
=>\(\dfrac{x}{y}\)=\(\dfrac{I}{III}\)=\(\dfrac{1}{3}\)=>\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy CTHH là Fe(NO3)3
PTK Fe(NO3)3=56+ (14+16.3).3=56+186=242 đvC
- Đặt CTHH dạng: \(Fe_x^{III}\left(PO_4\right)_y^{III}\)
Ta có: III.x=III.y
=>\(\dfrac{x}{y}\)=\(\dfrac{III}{III}\)=\(\dfrac{3}{3}\)=>\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy CTHH là FePO4
PTK FePO4=56+31+16.4 =56+31+64=151 đvC
- Đặt CTHH dạng: \(Fe_x^{III}OH_Y^I\)
Ta có: III.x=I.y
=>\(\dfrac{x}{y}\)=\(\dfrac{I}{III}\)=\(\dfrac{1}{3}\)=>\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy CTHH là FeOH3
PTK FeOH3=56+16+1.3 =75 đvC