Đáp án:
\({C_{M{\text{ }}{{\text{K}}_2}C{O_3}}} = 0,6M\)
\({C_{M{\text{ KOH}}}} = 1,4M\)
Giải thích các bước giải:
Ta có:
\({n_{C{O_2}}} = \frac{{10,56}}{{44}} = 0,24{\text{ mol;}}{{\text{n}}_{KOH}} = 0,4.2,6 = 1,04{\text{ mol}}\)
\( \to \frac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \frac{{1,04}}{{0,24}} > 2\) do vậy \(KOH\) dư
\(2KOH + C{O_2}\xrightarrow{{}}{K_2}C{O_3} + {H_2}O\)
Ta có:
\({n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,24{\text{ mol;}}{{\text{n}}_{KOH{\text{ dư}}}} = 1,04 - 0,24.2 = 0,56{\text{ mol}}\)
\({V_{dd}} = 400{\text{ ml = 0}}{\text{,4 lít}}\)
\( \to {C_{M{\text{ }}{{\text{K}}_2}C{O_3}}} = \frac{{0,24}}{{0,4}} = 0,6M\)
\({C_{M{\text{ KOH}}}} = \frac{{0,56}}{{0,4}} = 1,4M\)
