Em tham khảo nha :
\(\begin{array}{l}
a)\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
{m_{{H_2}O}} = {m_{{\rm{dd}}}} - {m_{{P_2}{O_5}}} = 400 - 11,36 = 388,64g\\
b)\\
{n_{{P_2}{O_5}}} = \dfrac{{11,36}}{{142}} = 0,08mol\\
{n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,16mol\\
{m_{{H_3}P{O_4}}} = 0,16 \times 98 = 15,68g\\
C{\% _{{H_3}P{O_4}}} = \dfrac{{15,68}}{{400}} \times 100\% = 3,92\%
\end{array}\)