Answer:
\[A=Borax(N{{a}_{2}}{{B}_{4}}{{O}_{7}})\]
\[X={{H}_{3}}B{{O}_{3}}\]
\[Z={{B}_{2}}{{O}_{3}}\]
\[\underset{\begin{smallmatrix}
(A) \\
Borax
\end{smallmatrix}}{\mathop{N{{a}_{2}}{{B}_{4}}{{O}_{7}}}}\,+2HCl+5{{H}_{2}}O\xrightarrow{{}}2NaCl+\underset{\begin{smallmatrix}
(X) \\
Boric\,acid
\end{smallmatrix}}{\mathop{4{{H}_{3}}B{{O}_{3}}}}\,\]
\[2{{H}_{3}}B{{O}_{3}}\xrightarrow[\begin{smallmatrix}
370K \\
(-2{{H}_{2}}O)
\end{smallmatrix}]{\Delta
}2HB{{O}_{2}}\xrightarrow[\begin{smallmatrix}
>370K \\
(-{{H}_{2}}O)
\end{smallmatrix}]{\Delta
}\underset{\begin{smallmatrix}
(Z) \\
Boric
\\
anlydrode
\end{smallmatrix}}{\mathop{{{B}_{2}}{{O}_{3}}}}\,\]
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