Question

what are the action of Boron on NaOH

Answer 1

B + 6NaOH > 2Na3BO3 + 3H2
product name is Sodium tetrahydroxoborate.

Answer 2

Answer: The products formed by the action of boron on NaOH are sodium borate and hydrogen gas.

Explanation:

The reaction of boron with sodium hydroxide leads to the formation of sodium borate and hydrogen gas.

The equation for the above reaction follows:

$2B+6NaOH\rightarrow 2Na_3BO_3+3H_2$

By Stoichiometry of the reaction:

2 moles of boron reacts with 6 moles of sodium hydroxide to produce 2 moles of sodium borate and 3 moles of hydrogen gas.