Đáp án:

\(C{\% _{N{a_2}HP{O_4}}} = \dfrac{{0,2 \times 142}}{{69,6}} \times 100\%  = 40,8\% \)

Giải thích các bước giải:

\(\begin{array}{l}
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
{n_{{P_2}{O_5}}} = 0,1mol\\
{n_{{H_2}O}} = 0,3mol\\
 \to {n_{{P_2}{O_5}}} = \dfrac{1}{3}{n_{{H_2}O}} = 0,1mol\\
 \to {n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,2mol\\
 \to {n_{NaOH}} = \dfrac{{50 \times 32}}{{100 \times 40}} = 0,4mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{{H_3}P{O_4}}}}} = 2
\end{array}\)

Suy ra chỉ tạo 1 muối \(N{a_2}HP{O_4}\)

\(\begin{array}{l}
2NaOH + {H_3}P{O_4} \to N{a_2}HP{O_4} + 2{H_2}O\\
 \to {n_{N{a_2}HP{O_4}}} = {n_{{H_3}P{O_4}}} = 0,2mol\\
{m_{{\rm{dd}}}} = 14,2 + 5,4 + 50 = 69,6g\\
 \to C{\% _{N{a_2}HP{O_4}}} = \dfrac{{0,2 \times 142}}{{69,6}} \times 100\%  = 40,8\% 
\end{array}\)

\(\begin{array}{l}
n_{P_2O_5}=\frac{14,2}{142}=0,1(mol)\\
n_{H_2O}=\frac{5,4}{18}=0,3(mol)\\
P_2O_5+3H_2O\to 2H_3PO_4\\
\to Phan\,ung\,hoan\,toan\\
Theo\,PT:\,n_{H_3PO_4}=2n_{P_2O_5}=0,2(mol)\\
n_{NaOH}=\frac{50.32\%}{40}=0,4(mol)\\
T=\frac{n_{NaOH}}{n_{H_3PO_4}}=\frac{0,4}{0,2}=2\\
\to Tao\,muoi\,Na_2HPO_4\\
H_3PO_4+2NaOH\to Na_2HPO_4+H_2O\\
Theo\,PT:\,n_{Na_2HPO_4}=n_{H_3PO_4}=0,2(mol)\\
\to C\%_{Na_2HPO_4}=\frac{0,2.142}{14,2+5,4+50}.100\%\approx 40,8\%
\end{array}\)