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Boron(III) iodide react with oxygen

2BI3 + 9O2 → B2O3 + 3I2O5




Boron(III) iodide react with oxygen to produce boron trioxide(III) and oxide iodine(V). This reaction takes place at a temperature of 150-175°C.


Picture of reaction: img1 BB html

Сoding to search: 2 BI3 + 9 O2 cnd [ temp ] = B2O3 + 3 I2O5

Add / Edited: 02.02.2015 /

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


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