Giải thích các bước giải:

1,

a, Dung dịch A thuộc hợp chất axit. Axit phosphoric

b,

\(\begin{array}{l}
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
{n_{{P_2}{O_5}}} = 0,025mol\\
{n_{{H_2}O}} = 13,41mol\\
 \to {n_{{P_2}{O_5}}} < {n_{{H_2}O}}
\end{array}\)

\({P_2}{O_5}\) dư

\(\begin{array}{l}
 \to {n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,05mol \to {m_{{H_3}P{O_4}}} = 4,9g\\
{m_{{\rm{dd}}}} = 245g\\
 \to C\% {H_3}P{O_4} = \dfrac{{4,9}}{{245}} \times 100\%  = 0,02\% 
\end{array}\)

CM bị thiếu dữ kiện nha bạn

2,

Bảo toàn nguyên tố H, ta có:

\(\begin{array}{l}
F{e_2}{O_3} + {H_2} \to 2Fe + 3{H_2}O\\
ZnO + {H_2} \to Zn + {H_2}O\\
{n_{{H_2}}} = \dfrac{{PV}}{{RT}} = \dfrac{{1 \times 12}}{{0,082 \times (273 + 25)}} = 0,49mol\\
 \to {n_{{H_2}O}} = {n_{{H_2}}} = 0,49mol \to {m_{{H_2}O}} = 8,82g
\end{array}\)

\(\begin{array}{l}
\dfrac{{{m_{F{e_2}{O_3}}}}}{{{m_{ZnO}}}} = \dfrac{{80}}{{81}} \to \dfrac{{{n_{F{e_2}{O_3}}}}}{{{n_{ZnO}}}} = \dfrac{1}{2} \to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{ZnO}}\\
{n_{{H_2}}} = {n_{hỗn hợp}} = 0,49mol\\
 \to {n_{F{e_2}{O_3}}} + {n_{ZnO}} = 0,49 \to \dfrac{1}{2}{n_{ZnO}} + {n_{ZnO}} = 0,49\\
 \to {n_{ZnO}} = 0,33mol \to {n_{F{e_2}{O_3}}} = 0,165mol\\
 \to {m_{ZnO}} = 26,73g \to {m_{F{e_2}{O_3}}} = 26,4g
\end{array}\)

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{HCl}} = 2({n_{Fe}} + {n_{Zn}}) = 2(2{n_{F{e_2}{O_3}}} + {n_{ZnO}}) = 1,98mol\\
 \to {m_{HCl}} = 72,27g \to {m_{HCl}}{\rm{dd = }}\dfrac{{72,27 \times 100}}{{18,25}} = 396g
\end{array}\)

3,

Bảo toàn nguyên tố H, ta có:

\(\begin{array}{l}
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{{H_2}}} = 0,65mol\\
 \to {n_{{H_2}O}} = {n_{{H_2}}} = 0,65mol \to {m_{{H_2}O}} = 11,7g\\
\dfrac{{{m_{F{e_3}{O_4}}}}}{{{m_{CuO}}}} = \dfrac{{21}}{{20}} \to \dfrac{{{n_{F{e_3}{O_4}}}}}{{{n_{CuO}}}} = \dfrac{{21}}{{58}} \to {n_{F{e_3}{O_4}}} = \dfrac{{21}}{{58}}{n_{CuO}}\\
{n_{{H_2}O}} = {n_{hỗn hợp}} = 0,65mol\\
 \to {n_{F{e_3}{O_4}}} + {n_{CuO}} = 0,65 \to \dfrac{{21}}{{58}}{n_{CuO}} + {n_{CuO}} = 0,65\\
 \to {n_{CuO}} = 0,48mol \to {n_{F{e_3}{O_4}}} = 0,17mol\\
 \to {m_{CuO}} = 38,4g \to {m_{F{e_3}{O_4}}} = 39,44g\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = 2{n_{Fe}} = 2 \times 3{n_{F{e_3}{O_4}}} = 1,02mol\\
 \to {m_{HCl}} = 37,23g \to {m_{HCl}}{\rm{dd = }}\dfrac{{37,23 \times 100}}{{18,25}} = 204g
\end{array}\)