CuI + H2SO4 + Fe2O3 = CuSO3 + FeSO4 + I2 + H2O |

2CuI + 6H₂SO₄ + 2Fe₂O₃ 2CuSO₃ + 4FeSO₄ + I₂ + 6H₂O

Copper(I) iodide react with sulfuric acid and iron(III) oxide to produce copper sulfate, iron(II) sulfate iodine and water. Sulfuric acid - concentrated solution. This reaction takes place at a temperature near 80°C.

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Copper(I) iodide react with sulfuric acid and iron(III) oxide

2CuI + 6H₂SO₄ + 2Fe₂O₃ 2CuSO₃ + 4FeSO₄ + I₂ + 6H₂O

Picture of reaction:

Сoding to search: 2 CuI + 6 H2SO4 + 2 Fe2O3 cnd [ temp ] = 2 CuSO3 + 4 FeSO4 + I2 + 6 H2O

Add / Edited: 08.04.2015 /

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

Copper(I) iodide react with sulfuric acid and iron(III) oxide

2CuI + 6H2SO4 + 2Fe2O3 2CuSO3 + 4FeSO4 + I2 + 6H2O

Picture of reaction:

Сoding to search: 2 CuI + 6 H2SO4 + 2 Fe2O3 cnd [ temp ] = 2 CuSO3 + 4 FeSO4 + I2 + 6 H2O

Add / Edited: 08.04.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

2CuI + 6H₂SO₄ + 2Fe₂O₃ 2CuSO₃ + 4FeSO₄ + I₂ + 6H₂O

Add / Edited: 08.04.2015 /