Đáp án:
\(\begin{array}{l}
C{\% _{NaOH}} = 20,145\% \\
C{\% _{N{a_3}P{O_4}}} = 24,6\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{P_2}{O_5} + 6NaOH \to 2N{a_3}P{O_4} + 3{H_2}O\\
{n_{{P_2}{O_5}}} = \dfrac{{21,3}}{{142}} = 0,15mol\\
{n_{NaOH}} = 6{n_{{P_2}{O_5}}} = 0,9mol\\
{m_{NaOH}} = 0,9 \times 40 = 36g\\
{n_{N{a_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,3mol\\
{m_{N{a_3}P{O_4}}} = 0,3 \times 164 = 49,2g\\
{m_{{\rm{dd}}spu}} = 21,3 + 178,7 = 200g\\
C{\% _{NaOH}} = \dfrac{{36}}{{178,7}} \times 100\% = 20,145\% \\
C{\% _{N{a_3}P{O_4}}} = \dfrac{{49,2}}{{200}} \times 100\% = 24,6\%
\end{array}\)
