Giải thích các bước giải:
\(\begin{array}{l}
a)2Fe + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Fe}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 19,6\% }}{{100\% }} = 39,2g\\
\to {n_{{H_2}S{O_4}}} = 0,4mol\\
\to {n_{{H_2}S{O_4}}} > {n_{Fe}} \to {n_{{H_2}S{O_4}}}dư\\
b)
\end{array}\)
Dung dịch sau phản ứng có: \(F{e_2}{(S{O_4})_3}\)và \({H_2}S{O_4}\) dư
\(\begin{array}{l}
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Fe}} = 0,1mol\\
\to {m_{F{e_2}{{(S{O_4})}_3}}} = 0,1 \times 400 = 40g\\
{n_{{H_2}S{O_4}(theopt)}} = \dfrac{3}{2}{n_{Fe}} = 0,3mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,4 - 0,3 = 0,1mol\\
\to {m_{{H_2}S{O_4}(dư)}} = 0,1 \times 98 = 9,8g\\
\to {n_{{H_2}}} = \dfrac{3}{2}{n_{Fe}} = 0,3mol\\
\to {m_{{H_2}}} = 0,6g\\
\to {m_{{\rm{dd}}}} = {m_{Fe}} + {m_{{\rm{dd}}{{\rm{H}}_2}S{O_4}}} - {m_{{H_2}}} = 11,2 + 200 - 0,6 = 210,6g\\
\to \% {m_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{40}}{{210,6}} \times 100\% = 19\% \\
\to \% {m_{{H_2}S{O_4}(du)}} = \dfrac{{9,8}}{{210,6}} \times 100\% = 4,65\% \\
{m_{{\rm{dd}}}} = V \times D \to V = \dfrac{{210,6}}{{1,05}} = 201ml = 0,201l\\
\to C{M_{F{e_2}{{(S{O_4})}_3}}} = C{M_{{H_2}S{O_4}(dư)}} = \dfrac{{0,1}}{{0,201}} = 0,5M
\end{array}\)
