BI3 + NH3 = B2(NH)3 + NH4I |

2BI₃ + 9NH₃ → B₂(NH)₃ + 6NH₄I

Boron(III) iodide react with ammonia to produce boron(III) imide and ammonium iodide. Ammonia - liquid. This reaction takes place at a temperature near -78°C.

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Boron(III) iodide react with ammonia

2BI₃ + 9NH₃ → B₂(NH)₃ + 6NH₄I

Picture of reaction:

Сoding to search: 2 BI3 + 9 NH3 = B2NH3 + 6 NH4I

Add / Edited: 02.02.2015 /

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Boron(III) iodide react with ammonia

2BI3 + 9NH3 → B2(NH)3 + 6NH4I

Picture of reaction:

Сoding to search: 2 BI3 + 9 NH3 = B2NH3 + 6 NH4I

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

2BI₃ + 9NH₃ → B₂(NH)₃ + 6NH₄I

Add / Edited: 02.02.2015 /