BI3 + H2SO4 + H2O = H3BO3 + I2 + H2S |

8BI₃ + 3H₂SO₄ + 12H₂O → 8H₃BO₃ + 12I₂ + 3H₂S

Boron(III) iodide react with sulfuric acid and water to produce orthoboric acid, iodine and hydrogen sulfide. Sulfuric acid - concentrated solution. The reaction takes place in a boiling solution.

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Boron(III) iodide react with sulfuric acid and water

8BI₃ + 3H₂SO₄ + 12H₂O → 8H₃BO₃ + 12I₂ + 3H₂S

Picture of reaction:

Сoding to search: 8 BI3 + 3 H2SO4 + 12 H2O = 8 H3BO3 + 12 I2 + 3 H2S

Add / Edited: 02.02.2015 /

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

Boron(III) iodide react with sulfuric acid and water

8BI3 + 3H2SO4 + 12H2O → 8H3BO3 + 12I2 + 3H2S

Picture of reaction:

Сoding to search: 8 BI3 + 3 H2SO4 + 12 H2O = 8 H3BO3 + 12 I2 + 3 H2S

Add / Edited: 02.02.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

8BI₃ + 3H₂SO₄ + 12H₂O → 8H₃BO₃ + 12I₂ + 3H₂S

Add / Edited: 02.02.2015 /