Đáp án:
\(\begin{array}{l}
a/\\
{V_{{H_2}}} = 25\% \\
{V_{{H_2}S}} = 75\% \\
b/C{M_{{H_2}S{O_4}}} = 0,412M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + S \to FeS\\
FeS + {H_2}S{O_4} \to FeS{O_4} + {H_2}S\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,2mol\\
{n_S} = 0,15mol
\end{array}\)
\(\begin{array}{l}
\to {n_S} < {n_{Fe}} \to {n_{Fe}}dư\\
\to {n_{Fe}} = {n_{FeS}} = {n_S} = 0,15mol\\
\to {n_{Fe(dư)}} = 0,05mol\\
\to {n_{{H_2}}} = {n_{Fe(dư)}} = 0,05mol\\
\to {n_{{H_2}S}} = {n_{FeS}} = 0,15mol
\end{array}\)
\(\begin{array}{l}
a/\\
{V_{{H_2}}} = \dfrac{{0,05 \times 22,4}}{{(0,05 + 0,15) \times 22,4}} \times 100\% = 25\% \\
{V_{{H_2}S}} = 100\% - 25\% = 75\%
\end{array}\)
\(\begin{array}{l}
b/\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,0125mol\\
\to {n_{{H_2}S{O_4}(dư)}} = \dfrac{1}{2}{n_{NaOH}} = 0,00625mol\\
{n_{{H_2}S{O_4}}} = {n_{Fe(dư)}} + {n_{FeS}} = 0,2mol\\
\to {n_{{H_2}S{O_4}}} = 0,00625 + 0,2 = 0,206mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,206}}{{0,5}} = 0,412M
\end{array}\)
