Đáp án:
\(\frac{a}{b} = 1,47778\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2}\)
\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
Ta có:
\({n_{KCl{O_3}}} = \frac{a}{{122,5}} = {n_{KCl}} \to {m_{KCl}} = \frac{a}{{122,5}}.74,5 = \frac{{74,5a}}{{122,5}}\)
\({n_{KMn{O_4}}} = \frac{b}{{158}} \to {n_{{K_2}Mn{O_4}}} = {n_{Mn{O_2}}} = \frac{1}{2}{n_{KMn{O_4}}} = \frac{b}{{316}} \to {m_{{K_2}Mn{O_4}}} + {n_{Mn{O_2}}} = \frac{b}{{316}}.(39.2 + 55 + 16.4 + 55 + 16.2) = \frac{{71b}}{{79}} = \frac{{74,5a}}{{122,5}} \to \frac{a}{b} = 1,47778\)
Ta có:
\({n_{{O_2}{\text{ trong }}{\text{KCl}}{{\text{O}}_3}}} = \frac{3}{2}.{n_{KCl{O_3}}} = \frac{3}{2}.\frac{{a}}{{122,5}};{n_{{O_2}{\text{ trong KMn}}{{\text{O}}_4}}} = \frac{1}{2}{n_{KMn{O_4}}} = \frac{b}{{316}}\)
Vì % số mol=% thể tích
\( \to \frac{{{V_{{O_2}{\text{ KCl}}{{\text{O}}_3}}}}}{{{V_{{O_2}{\text{KMn}}{{\text{O}}_4}}}}} = \frac{{\frac{3}{2}.\frac{a}{{122,5}}}}{{\frac{b}{{316}}}} = 6,879\)
