Đáp án:

b) \({C_{M{\text{ KI}}}} = 0,4M\)

c) \({V_{{O_3}}} = 8,96{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2KI + {O_3} + {H_2}O\xrightarrow{{}}2KOH + {O_2} + {I_2}\)

Ta có: 

\({V_B} = {V_{{N_2}}} + {V_{{O_3}}} = 20{\text{ lít}}\)

\({I_2} + 2N{a_2}{S_2}{O_3}\xrightarrow{{}}2NaI + N{a_2}{S_4}{O_6}\)

Ta có: 

\({n_{N{a_2}{S_2}{O_3}}} = 4.0,2 = 0,8{\text{ mol}} \to {{\text{n}}_{{I_2}}} = \frac{1}{2}{n_{N{a_2}{S_2}{O_3}}} = 0,4{\text{ mol}} \to {{\text{n}}_{{O_3}}} = {n_{{I_2}}} = 0,4{\text{ mol;}}{{\text{n}}_{KI}} = 2{n_{{I_2}}} = 0,8{\text{ mol}}\)

\( \to {C_{M{\text{ KI}}}} = \frac{{0,8}}{2} = 0,4M\)

\({V_{{O_3}}} = 0,4.22,4 = 8,96{\text{ lít}} \to {{\text{V}}_{{N_2}}} = 20 - 8,96 = 11,04{\text{ lít}}\)

\({n_{{N_2}}} = \frac{{11,04}}{{22,4}} = 0,49286{\text{ mol}} \to {{\text{m}}_B} = {m_{{O_2}}} + {m_{{N_2}}} = 0,4.32 + 0,49286.28 = 26,6{\text{ gam;}}{{\text{n}}_B} = 0,4 + 0,49286 = 0,89286{\text{ mol}}\)

\( \to {M_B} = \frac{{26,6}}{{0,89286}} = 29,8 \to {d_{B/He}} = \frac{{29,8}}{4} = 7,45;{d_{B/kk}} = \frac{{29,8}}{{29}} = 1,075\)