Đáp án:

 $m = 83,3 gam$

Giải thích các bước giải:

$\begin{gathered}   {P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4} \hfill \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 142\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2.98 = 196\,\,\,\,\,\,\,\,\,\,gam \hfill \\  \end{gathered} $

Coi $P_2O_5$ là dung dịch $H_3PO_4$ có: $C\%  = \frac{{196}}{{142}} \cdot 100\%  = 138\% $

Theo phương pháp đường chéo, ta có:

\[\begin{gathered}   {P_2}{O_5}\,\,\,\,\,138\% \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15\%  \hfill \\   \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,50\%  \hfill \\   {H_3}P{O_4}\,\,\,\,\,35\% \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,88\%  \hfill \\    \to \frac{{{m_{{P_2}{O_5}}}}}{{{m_{{H_3}P{O_4}}}}} = \frac{{15}}{{88}} \to {m_{{H_3}P{O_4}}} = \frac{{14,2.88}}{{15}} = 83,3\,\,gam \hfill \\  \end{gathered} \]