$Na_2B_4O_7 \stackrel{\text{heat}}{\longrightarrow} X + NaBO_2$
In the above reaction,the product $X$ is:

  • A
    $H_3BO_3$
  • B
    $B_2O_3$
  • C
    $Na_2B_2O_5$
  • D
    $NaB_3O_5$

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