1)

Ta có:

\({n_{Ca{{(OH)}_2}}} = 0,1.1,25 = 0,125{\text{ mol;}}{{\text{n}}_{S{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)

\( \to \frac{{{n_{S{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \frac{{0,15}}{{0,125}} = 1,2\)

Vậy phản ứng tạo ra 2 muối:

\(Ca{(OH)_2} + S{O_2}\xrightarrow{{}}CaS{O_3} + {H_2}O\)

\(Ca{(OH)_2} + 2{\text{S}}{O_2}\xrightarrow{{}}Ca{(H{\text{S}}{O_3})_2}\)

\( \to {n_{Ca{{(H{\text{S}}{O_3})}_2}}} = {n_{S{O_2}}} - {n_{Ca{{(OH)}_2}}} = 0,15 - 0,125 = 0,025{\text{ mol}}\)

\( \to {C_{M{\text{ Ca(HS}}{O_3}{)_2}}} = \frac{{0,025}}{{0,1}} = 0,25M\)

Chọn \(B\)

2)

Ta có:

\({n_{{H_2}S}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,4.2 = 0,8{\text{ mol}}\)

\( \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,8}}{{0,25}} = 3,2 > 2\)

Vậy \(NaOH\) dư

Phản ứng xảy ra:

\(2NaOH + {H_2}S\xrightarrow{{}}N{a_2}S + 2{H_2}O\)

\( \to {n_{N{a_2}S}} = {n_{{H_2}S}} = 0,25{\text{ mol;}}{{\text{n}}_{NaOH{\text{ dư}}}} = 0,8 - 0,25 = 0,3{\text{ mol}}\)

\( \to {C_{M{\text{ N}}{{\text{a}}_2}S}} = \frac{{0,25}}{{0,4}} = 0,625M;{C_{M{\text{ NaOH}}}} = \frac{{0,3}}{{0,4}} = 0,75M\)

Chọn \(A\)

3)

Ta có:

\({n_{{H_2}S}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol;}}{{\text{m}}_{KOH}} = 200.11,2\%  = 22,4{\text{ gam}}\)

\( \to {n_{KOH}} = \frac{{22,4}}{{56}} = 0,4{\text{ mol}} \to \frac{{{n_{KOH}}}}{{{n_{{H_2}S}}}} = \frac{{0,4}}{{0,25}} = 1,6\)

Vậy phản ứng tạo ra 2 muối

Phản ứng xảy ra:

\(2KOH + {H_2}S\xrightarrow{{}}{K_2}S + 2{H_2}O\)

\(KOH + {H_2}S\xrightarrow{{}}KH{\text{S}} + {H_2}O\)

\( \to {n_{{K_2}S}} = {n_{KOH}} - {n_{{H_2}S}} = 0,4 - 0,25 = 0,15{\text{ mol}}\)

\( \to {m_{{K_2}S}} = 0,15.(39.2 + 32) = 16,5{\text{ gam}}\)

\( \to {m_{dd}} = {m_{dd\;{\text{KOH}}}} + {m_{{H_2}S}} = 200 + 0,25.34 = 208,5{\text{ gam}}\)

\( \to C{\% _{{K_2}S}} = \frac{{16,5}}{{208,5}} = 7,91\% \)

Chọn \(D\)