Đáp án:
\(\begin{array}{l}
a)\% {V_{{H_2}S}} = \% {V_{{H_2}}} = 50\% \\
b)C{\% _{N{a_2}S{O_3}}} = 60,5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + S \to FeS\\
FeS + 2HCl \to FeC{l_2} + {H_2}S\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 0,1mol\\
{n_S} = 0,05mol\\
\to {n_{Fe}} > {n_S} \to {n_{Fe}}dư
\end{array}\)
\(\begin{array}{l}
{n_{Fe}} = {n_{FeS}} = {n_S} = 0,05mol\\
\to {n_{Fe(dư)}} = 0,05mol\\
a)\\
{n_{{H_2}S}} = {n_{FeS}} = 0,05mol\\
{n_{{H_2}}} = {n_{Fe(dư)}} = 0,05mol
\end{array}\)
\( \to \% {V_{{H_2}S}} = \% {V_{{H_2}}} = \dfrac{{0,05 \times 22,4}}{{0,05 \times 22,4 + 0,05 \times 22,4}} \times 100\% = 50\% \)
\(\begin{array}{l}
b)\\
2{H_2}S + 3{O_2} \to 2S{O_2} + 2{H_2}O2\\
{n_{S{O_2}}} = {n_{{H_2}S}} = 0,05mol\\
{m_{NaOH}} = \dfrac{{100 \times 20\% }}{{100\% }} = 20g\\
\to {n_{NaOH}} = 0,5mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,5}}{{0,05}} = 10
\end{array}\)
=> Tạo 1 muối: \(N{a_2}S{O_3}\)
\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,05mol\\
\to {m_{N{a_2}S{O_3}}} = 62,3g\\
{m_{{\rm{dd}}}} = {m_{S{O_2}}} + {m_{{\rm{dd}}NaOH}} = 103g\\
\to C{\% _{N{a_2}S{O_3}}} = \dfrac{{62,3}}{{103}} \times 100\% = 60,5\%
\end{array}\)
