${\text{n}}_{\text{NaOH}}=0,4 \text{mol};{\text{n}}_{{\text{H}}_3{\text{PO}}_4}=0,15 \text{mol}$

$\frac{{\text{n}}_{\text{NaOH}}}{{\text{n}}_{{\text{H}}_3{\text{PO}}_4}}=\frac{0,4}{0,15}=2,67\to$ Muối thu được gồm: Na₂HPO₄ (x mol) và Na₃PO₄ (y mol)

$$\begin{gathered} \begin{array}{l}2\text{NaOH}+{\text{H}}_3{\text{PO}}_4\to {\text{Na}}_2{\text{HPO}}_4+2{\text{H}}_2\text{O}\\ 2\text{x\hspace{0.17em}mol\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}mol\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}mol}\end{array} \\ \begin{array}{l}3\text{NaOH}+{\text{H}}_3{\text{PO}}_4\to {\text{Na}}_3{\text{PO}}_4+3{\text{H}}_2\text{O}\\ 3\text{y\hspace{0.17em}mol\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}mol\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}mol}\end{array} \\ \left\{\begin{array}{l}x+y=0,15\\ 2x+3y=0,4\end{array}\to \left\{\begin{array}{l}x=0,05\\ y=0,1\end{array}\right.\right. \end{gathered}$$

Sau khi các phản ứng xảy ra hoàn toàn, khối lượng muối thu được là:

$\text{m}=0,05.142+0,1.164=23,5\text{ gam. }$