Fe3I8 + (NH4)2CO3 + H2O = FeO(OH) + Fe(OH)2 + CO2 + NH4I |

Fe₃I₈ + 4(NH₄)₂CO₃ + 2H₂O → 2FeO(OH) + Fe(OH)₂ + 4CO₂ + 8NH₄I

Iron(II)-diiron(III) iodide react with ammonium carbonate and water to produce iron(III) oxide-hydroxide, iron(II) hydroxide, carbon dioxide and ammonium iodide.

Find another reaction

Thermodynamic properties of substances

The solubility of the substances

Periodic table of elements

Iron(II)-diiron(III) iodide react with ammonium carbonate and water

Fe₃I₈ + 4(NH₄)₂CO₃ + 2H₂O → 2FeO(OH) + Fe(OH)₂ + 4CO₂ + 8NH₄I

Picture of reaction:

Сoding to search: Fe3I8 + 4 NH42CO3 + 2 H2O = 2 FeOOH + FeOH2 + 4 CO2 + 8 NH4I

Add / Edited: 18.06.2015 /

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

Iron(II)-diiron(III) iodide react with ammonium carbonate and water

Fe3I8 + 4(NH4)2CO3 + 2H2O → 2FeO(OH) + Fe(OH)2 + 4CO2 + 8NH4I

Picture of reaction:

Сoding to search: Fe3I8 + 4 NH42CO3 + 2 H2O = 2 FeOOH + FeOH2 + 4 CO2 + 8 NH4I

Add / Edited: 18.06.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

Fe₃I₈ + 4(NH₄)₂CO₃ + 2H₂O → 2FeO(OH) + Fe(OH)₂ + 4CO₂ + 8NH₄I

Add / Edited: 18.06.2015 /