Giải thích các bước giải:
\(\begin{array}{l}
TH1:\\
4{H_3}P{O_4} + {P_2}{O_5} \to 3{H_4}{P_2}{O_7}\\
{m_{{\rm{dd}}{H_3}P{O_4}}} = 51,5g\\
\to {m_{{H_3}P{O_4}}} = \dfrac{{51,5 \times 5,7\% }}{{100\% }} = 2,94g\\
\to {n_{{H_3}P{O_4}}} = 0,03mol\\
{n_{{P_2}{O_5}}} = 0,06mol\\
\to {n_{{P_2}{O_5}}} > {n_{{H_3}P{O_4}}} \to {n_{{P_2}{O_5}}}dư\\
\to {n_{{H_4}{P_2}{O_7}}} = \dfrac{3}{4}{n_{{H_3}P{O_4}}} = 0,0225mol\\
\to {m_{{H_4}{P_2}{O_7}}} = 4,005g\\
{m_{{\rm{dd}}}} = {m_{{P_2}{O_5}}} + {m_{{\rm{dd}}{H_3}P{O_4}}} = 60,02g\\
\to C{\% _{{H_4}{P_2}{O_7}}} = \dfrac{{4,005}}{{60,02}} \times 100\% = 6,67\% \\
TH2:\\
3Ba{(OH)_2} + 2{H_3}P{O_4} \to B{a_3}{(P{O_4})_2} + 6{H_2}O\\
3KOH + {H_3}P{O_4} \to {K_3}P{O_4} + 3{H_2}O\\
{n_{{H_3}P{O_4}}} = 0,7mol\\
{n_{Ba{{(OH)}_2}}} = 0,3mol\\
{n_{KOH}} = 0,2mol\\
{n_{{H_3}P{O_4}}}(pt) = \dfrac{2}{3}{n_{Ba{{(OH)}_2}}} + \dfrac{1}{3}{n_{KOH}} = 0,267mol\\
\to {n_{{H_3}P{O_4}}}(dư) = 0,433mol\\
\to {m_{{H_3}P{O_4}}}(dư) = 42,4g\\
{n_{B{a_3}{{(P{O_4})}_2}}} = \dfrac{1}{3}{n_{Ba{{(OH)}_2}}} = 0,1mol\\
\to {m_{B{a_3}{{(P{O_4})}_2}}} = 60,1g\\
{n_{{K_3}P{O_4}}} = \dfrac{1}{3}{n_{KOH}} = 0,067mol\\
\to {m_{{K_3}P{O_4}}} = 14,2g\\
\to {m_{{\rm{dd}}}} = {m_{Ba{{(OH)}_2}}} + {m_{KOH}} + {m_{{H_3}P{O_4}}} = 51,3 + 11,2 + 68,6 = 131,1g\\
\to C{\% _{B{a_3}{{(P{O_4})}_2}}} = \dfrac{{60,1}}{{131,1}} \times 100\% = 45,84\% \\
\to C{\% _{{K_3}P{O_4}}} = \dfrac{{14,2}}{{131,1}} \times 100\% = 10,83\% \\
\to C{\% _{{H_3}P{O_4}}}(du) = \dfrac{{42,4}}{{131,1}} \times 100\% = 32,34\%
\end{array}\)
