Đáp án:
\( {C_{M{\text{ }}{{\text{K}}_2}S{O_3}}} = {C_{M{\text{ KOH}}}} = 0,5M\)
Giải thích các bước giải:
Ta có:
\({n_{S{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol;}}{{\text{n}}_{KOH}} = 0,2.1,5 = 0,3{\text{ mol}}\)
Ta có:
\(\frac{{{n_{KOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,3}}{{0,1}} = 3 > 2\)
Do vậy \(KOH\) dư.
\( \to {n_{{K_2}S{O_3}}} = {n_{S{O_2}}} = 0,1{\text{ mol}} \to {{\text{n}}_{KOH{\text{ dư}}}} = 0,3 - 0,1.2 = 0,1{\text{ mol}}\)
Ta có:
\({V_{dd\;{\text{X}}}} = 200{\text{ ml = 0}}{\text{,2 lít}}\)
\( \to {C_{M{\text{ }}{{\text{K}}_2}S{O_3}}} = {C_{M{\text{ KOH}}}} = \frac{{0,1}}{{0,2}} = 0,5M\)
