Đáp án:
a,
\(\begin{array}{l}
\% {m_{Fe}} = \dfrac{{8,4}}{{40,4}} \times 100\% = 20,79\% \\
\% {m_{F{e_2}{O_3}}} = 79,21\%
\end{array}\)
b,
\(C{M_{{H_2}S{O_4}}} = \dfrac{n}{V} = \dfrac{{0,75}}{{0,3}} = 2,5M\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,15mol \to {m_{Fe}} = 8,4g\\
{m_{F{e_2}{O_3}}} = 40,4 - 8,4 = 32g \to {n_{F{e_2}{O_3}}} = 0,2mol\\
\% {m_{Fe}} = \dfrac{{8,4}}{{40,4}} \times 100\% = 20,79\% \\
\% {m_{F{e_2}{O_3}}} = 79,21\% \\
{n_{{H_2}S{O_4}}} = {n_{Fe}} + 3{n_{F{e_2}{O_3}}} = 0,75mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{n}{V} = \dfrac{{0,75}}{{0,3}} = 2,5M
\end{array}\)
