Đáp án:
\({{\text{m}}_{NaHS}} = 1,12{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{{H_2}S}} = \frac{{7,168}}{{22,4}} = 0,32{\text{ mol; }}{{\text{m}}_{NaOH}} = 8.10\% = 0,8{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{{0,8}}{{40}} = 0,02{\text{ mol}}\)
\(\frac{{{n_{NaOH}}}}{{{n_{{H_2}S}}}} = \frac{{0,02}}{{0,32}} < 1\)
Vậy \({H_2}S\) dư.
\(NaOH + {H_2}S\xrightarrow{{}}NaHS + {H_2}O\)
\({n_{NaHS}} = {n_{NaOH}} = 0,02{\text{ mol}} \to {{\text{m}}_{NaHS}} = 0,02.(23 + 1 + 32) = 1,12{\text{ gam}}\)
