Đáp án:
\({C_{M{\text{ N}}{{\text{a}}_2}S{O_3}}} = 2,67;{C_{M{\text{ NaOH dư}}}} = 1,067M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(S + {O_2}\xrightarrow{{{t^o}}}S{O_2}\)
Ta có:
\({n_S} = \frac{{12,8}}{{32}} = 0,4{\text{ mol = }}{{\text{n}}_{S{O_2}}}\)
Ta có:
\({m_{dd{\text{ }}NaOH}} = 150.1,28 = 192{\text{ gam}} \to {{\text{m}}_{NaOH}} = 192.20\% = 38,4{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{{38,4}}{{40}} = 0,96{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,96}}{{0,4}} > 2\)
Vậy NaOH dư.
\( \to {n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,4{\text{ mol}}{{\text{n}}_{NaOH{\text{ dư}}}} = 0,96 - 0,4.2 = 0,16{\text{ mol}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}S{O_3}}} = \frac{{0,4}}{{0,15}} = 2,67;{C_{M{\text{ NaOH dư}}}} = \frac{{0,16}}{{0,15}} = 1,067M\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = 192 + 0,4.64 = 217,6{\text{ gam}}\)
\( \to {m_{N{a_2}S{O_3}}} = 0,4.(23.2 + 80) = 50,4{\text{ gam;}}{{\text{m}}_{NaOH{\text{ dư}}}} = 0,16.40 = 6,4{\text{ gam}}\)
\( \to C{\% _{N{a_2}S{O_3}}} = \frac{{50,4}}{{217,6}} = 23,16\% ;C{\% _{NaOH{\text{ dư}}}} = \frac{{6,4}}{{217,6}} = 2,94\% \)
