Đáp án:
\( {C_{M{\text{ NaOH}}}} = {C_{M{\text{ N}}{{\text{a}}_2}S}} = \frac{1}{3}M\)
\({m_{NaOH}} = \frac{{40}}{3}{\text{ gam;}}{{\text{m}}_{N{a_2}S}} = 26{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{NaOH}} = 0,3.1 = 0,3{\text{ mol;}}{{\text{n}}_{{H_2}S}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
Ta có:
\(\frac{{{n_{NaOH}}}}{{{n_{{H_2}S}}}} = \frac{{0,3}}{{0,1}} = 3 > 2\)
Vậy \(NaOH\) dư.
\( \to {n_{N{a_2}S}} = {n_{{H_2}S}} = 0,1{\text{ mol;}}{{\text{n}}_{NaOH{\text{ dư}}}} = 0,3 - 0,1.2 = 0,1{\text{ mol}}\)
\( \to {V_{dd}} = 300{\text{ ml = 0}}{\text{,3 lít}}\)
\( \to {C_{M{\text{ NaOH}}}} = {C_{M{\text{ N}}{{\text{a}}_2}S}} = \frac{{0,1}}{{0,3}} = \frac{1}{3}M\)
\({m_{NaOH}} = 40.\frac{1}{3} = \frac{{40}}{3}{\text{ gam;}}{{\text{m}}_{N{a_2}S}} = \frac{1}{3}.(23.2 + 32) = 26{\text{ gam}}\)
