LiI + H2SO4 = I2 + H2S + Li2SO4 + H2O |

8LiI + 5H₂SO₄ → 4I₂ + H₂S + 4Li₂SO₄ + 4H₂O

Lithium iodide react with sulfuric acid to produce iodine, hydrogen sulfide, lithium sulfate and water. Lithium iodide - solid. Sulfuric acid - concentrated solution. The reaction takes place in a boiling solution.

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Lithium iodide react with sulfuric acid

8LiI + 5H₂SO₄ → 4I₂ + H₂S + 4Li₂SO₄ + 4H₂O

Picture of reaction:

Сoding to search: 8 LiI + 5 H2SO4 = 4 I2 + H2S + 4 Li2SO4 + 4 H2O

Add / Edited: 12.06.2015 /

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

Lithium iodide react with sulfuric acid

8LiI + 5H2SO4 → 4I2 + H2S + 4Li2SO4 + 4H2O

Picture of reaction:

Сoding to search: 8 LiI + 5 H2SO4 = 4 I2 + H2S + 4 Li2SO4 + 4 H2O

Add / Edited: 12.06.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

8LiI + 5H₂SO₄ → 4I₂ + H₂S + 4Li₂SO₄ + 4H₂O

Add / Edited: 12.06.2015 /