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Iron(II) iodide react with zinc and phosphorus(III) fluoride. This reaction takes place at a temperature near 130°C and an overpressure

FeI2 + Zn + 5PF3 → [Fe(PF3)5] + ZnI2




Iron(II) iodide react with zinc and phosphorus(III) fluoride. This reaction takes place at a temperature near 130°C and an overpressure.


Picture of reaction: img1 BB html

Сoding to search: FeI2 + Zn + 5 PF3 cnd [ temp ] = FePF35 + ZnI2

Add / Edited: 19.04.2015 /

Evaluation of information: 5.0 out of 5 / number of votes: 1


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