Đáp án:
$15,6{\text{ gam}}$
Giải thích các bước giải:
Ta có:
\({n_{{H_2}S}} = \dfrac{{4,48}}{{22,4}} = 0,2{\text{ mol;}}\\{{\text{n}}_{NaOH}} = 0,25.2 = 0,5{\text{ mol}}\)
\( \to \dfrac{{{n_{NaOH}}}}{{{n_{{H_2}S}}}} = \dfrac{{0,5}}{{0,2}} > 2\)
Vậy \(NaOH\) dư
\(2NaOH + {H_2}S\xrightarrow{{}}N{a_2}S + 2{H_2}O\)
\( \to {n_{N{a_2}S}} = {n_{{H_2}S}} = 0,2{\text{ mol}}\)
\( \to {m_{N{a_2}S}} = 0,2.(23.2 + 32) = 15,6{\text{ gam}}\)