Giải thích các bước giải:
1,
\(\begin{array}{l}
{n_{{H_2}S}} = 0,04mol\\
{n_{KOH}} = 0,3mol\\
\to \dfrac{{{n_{KOH}}}}{{{n_{{H_2}S}}}} = 7,5
\end{array}\)
Tạo ra muối trung hòa, KOH dư
\(\begin{array}{l}
2KOH + {H_2}S \to {K_2}S + 2{H_2}O\\
\to {n_{{K_2}S}} = {n_{{H_2}S}} = 0,04mol\\
\to {m_{{K_2}S}} = 4,4g
\end{array}\)
2,
\(\begin{array}{l}
{n_{{H_2}S}} = 0,2mol\\
{n_{NaOH}} = 0,4mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{{H_2}S}}}} = 2
\end{array}\)
Tạo ra muối trung hòa
\(\begin{array}{l}
2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O\\
\to {n_{N{a_2}S}} = {n_{{H_2}S}} = 0,2mol\\
\to {m_{N{a_2}S}} = 15,6g\\
C{M_{N{a_2}S}} = \dfrac{{0,2}}{{0,2}} = 1M
\end{array}\)