Na3[IrCl6] + H2SO4 = Ir2(SO4)3 + Na2SO4 + HCl |

2Na₃[IrCl₆] + 6H₂SO₄ → Ir₂(SO₄)₃ + 3Na₂SO₄ + 12HCl

Sodium hexachloridoiridate(III) react with sulfuric acid to produce iridium(III) sulfate, sodium sulfate and hydrogen chloride. Sulfuric acid concentration less 40%. The reaction takes place in a boiling solution.

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Sodium hexachloridoiridate(III) react with sulfuric acid

2Na₃[IrCl₆] + 6H₂SO₄ → Ir₂(SO₄)₃ + 3Na₂SO₄ + 12HCl

Picture of reaction:

Сoding to search: 2 Na3IrCl6 + 6 H2SO4 = Ir2SO43 + 3 Na2SO4 + 12 HCl

Add / Edited: 31.05.2015 /

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

Sodium hexachloridoiridate(III) react with sulfuric acid

2Na3[IrCl6] + 6H2SO4 → Ir2(SO4)3 + 3Na2SO4 + 12HCl

Picture of reaction:

Сoding to search: 2 Na3IrCl6 + 6 H2SO4 = Ir2SO43 + 3 Na2SO4 + 12 HCl

Add / Edited: 31.05.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

2Na₃[IrCl₆] + 6H₂SO₄ → Ir₂(SO₄)₃ + 3Na₂SO₄ + 12HCl

Add / Edited: 31.05.2015 /