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Boron(III) iodide react with ammonia

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.

Picture of reaction: img1 BB html

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

Add / Edited: 02.02.2015 /

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

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