Đáp án:

14,

\(\% {m_{FeS}} = \) 94,02% và \(\% {m_{Fe}} = 5,98\% \)

15,

\({V_{hỗn hợp}} = 2,464l\)

Giải thích các bước giải:

14,

\(\begin{array}{l}
FeS + 2HCl \to FeC{l_2} + {H_2}S\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Pb{(N{O_3})_2} + {H_2}S \to PbS + 2HN{O_3}\\
{n_{hỗn hợp}} = 0,11mol\\
{n_{PbS}} = 0,1mol\\
 \to {n_{{H_2}S}} = {n_{PbS}} = 0,1mol\\
{n_{{H_2}S}} + {n_{{H_2}}} = 0,11 \to {n_{{H_2}}} = 0,01mol\\
 \to {n_{FeS}} = {n_{{H_2}S}} = 0,1mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,01mol\\
 \to \% {m_{FeS}} = \dfrac{{0,1 \times 88}}{{0,1 \times 88 + 0,01 \times 56}} \times 100\%  = 94,02\% \\
 \to \% {m_{Fe}} = 5,98\% 
\end{array}\)

15,

\(\begin{array}{l}
ZnS + 2HCl \to ZnC{l_2} + {H_2}S\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Pb{(N{O_3})_2} + {H_2}S \to PbS + 2HN{O_3}\\
{n_{PbS}} = 0,01mol\\
 \to {n_{{H_2}S}} = {n_{PbS}} = 0,01mol\\
 \to {n_{ZnS}} = {n_{{H_2}S}} = 0,01mol\\
 \to {m_{ZnS}} = 0,97g\\
 \to {m_{Zn}} = 7,47 - 0,97 = 6,5g\\
 \to {n_{Zn}} = 0,1mol\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
 \to {n_{{H_2}}} + {n_{{H_2}S}} = 0,1 + 0,01 = 0,11mol\\
 \to {V_{hỗn hợp}} = 2,464l
\end{array}\)