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Boron(III) fluoride react with sodium hydroxide

16BF3 + 14NaOH → 12Na[BF4] + Na2B4O7 + 7H2O




Boron(III) fluoride react with sodium hydroxide to produce sodium tetrafluoroborate(III), sodium tetraborate and water. Sodium hydroxide - diluted cold solution.


Picture of reaction: img1 BB html

Сoding to search: 16 BF3 + 14 NaOH = 12 NaBF4 + Na2B4O7 + 7 H2O

Add / Edited: 26.01.2015 /

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


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