How do you balance B_2Br_6 + HNO_3 -> B(NO_3)_3 + HBr?

Apr 20, 2017

For a start, your starting material is $B B {r}_{3}$.........

Explanation:

And another thing, while ""^(-)B(NO_3)_4 is known, I am pretty sure that $B {\left(N {O}_{3}\right)}_{3}$ is UNKNOWN. There was something about its synthesis in the old Russian literature, but given the source, its existence would be fairly suspect........

As an hypothetical reaction, we could write........

$B B {r}_{3} + 3 H N {O}_{3} + 3 N E {t}_{3} \rightarrow B {\left(N {O}_{3}\right)}_{3} + 3 E {t}_{3} N \cdot H B r$

Note that $B B {r}_{3}$ reacts very violently with water, and its aqueous chemistry is unknown.