Đáp án:
\(\begin{array}{l}
{C_\% }{H_2}S{O_4} \text{ dư } = 15,3125\% \\
{C_\% }F{e_2}{(S{O_4})_3} = 26,79\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{F{e_2}{O_3}}} = \dfrac{{24}}{{160}} = 0,15\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 39,2\% }}{{98}} = 0,8\,mol\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
\dfrac{{{n_{F{e_2}{O_3}}}}}{1} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,8 - 0,15 \times 3 = 0,35\,mol\\
{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,15\,mol\\
{m_{{\rm{dd}}spu}} = 24 + 200 = 224g\\
{C_\% }{H_2}S{O_4} \text{ dư } = \dfrac{{0,35 \times 98}}{{224}} \times 100\% = 15,3125\% \\
{C_\% }F{e_2}{(S{O_4})_3} = \dfrac{{0,15 \times 400}}{{224}} \times 100\% = 26,79\%
\end{array}\)
