Đáp án:
\(\begin{array}{l}
V = 0,4\\
{C_M}N{a_2}S = {C_M}NaHS = 0,33M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
ZnS + 2HCl \to ZnC{l_2} + {H_2}S\\
ZnC{l_2} + 2AgN{O_3} \to 2AgCl + Zn{(N{O_3})_2}\\
{n_{ZnS}} = \dfrac{{19,4}}{{97}} = 0,2\,mol\\
{n_{ZnC{l_2}}} = {n_{ZnS}} = 0,2\,mol\\
{n_{AgN{O_3}}} = 2{n_{ZnC{l_2}}} = 0,4\,mol\\
{V_{{\rm{dd}}AgN{O_3}}} = \dfrac{{0,4}}{1} = 0,4l \Rightarrow V = 0,4\\
{n_{NaOH}} = 0,3 \times 1 = 0,3\,mol\\
{n_{{H_2}S}} = {n_{ZnS}} = 0,2\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{{H_2}S}}}} = \dfrac{{0,3}}{{0,2}} = 1,5\\
1 < T < 2 \Rightarrow \text{ Tạo cả 2 muối } \\
2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O(1)\\
N{a_2}S + {H_2}S \to 2NaHS(2)\\
{n_{N{a_2}S(1)}} = {n_{{H_2}S(1)}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{n_{N{a_2}S(2)}} = {n_{{H_2}S(2)}} = 0,2 - 0,15 = 0,05\,mol\\
{n_{N{a_2}S}} = 0,15 - 0,05 = 0,1\,mol\\
{n_{NaHS}} = 0,05 \times 2 = 0,1\,mol\\
{C_M}N{a_2}S = {C_M}NaHS = \dfrac{{0,1}}{{0,3}} = 0,33M
\end{array}\)
