- Chọn B.

- Ta có:  ${{n}_{{{P}_{2}}{{O}_{5}}}}=\frac{28,4}{142}=0,2\left( mol \right);{{n}_{{{H}_{2}}O}}=\frac{90}{18}=5\left( mol \right)$

              ${{P}_{2}}{{O}_{5}}+3{{H}_{2}}O\to 2{{H}_{3}}P{{O}_{4}}$

       $0,2\left( mol \right)$        $5\left( mol \right)$

      $0,2\left( mol \right)$ $\to 0,6\left( mol \right)$           $0,4\left( mol \right)$

- ${{m}_{{{H}_{3}}P{{O}_{4}}}}=0,4\times 98=39,2\left( gam \right).$