Đáp án:
\(V=10,08 lít\)
\( \to \% {m_{KMn{O_4}}} = 58,8\% ;\% {m_{KCl{O_3}}} = 30,4\% ;\% {m_{Mn{O_2}}} = 10,8\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
Ta có:
\({n_{KCl}} = \frac{{14,9}}{{39 + 35,5}} = 0,2{\text{ mol = }}{{\text{n}}_{KCl{O_3}}}\)
\({m_Y} = \frac{{14,9}}{{22,508\% }} = 66,2{\text{ gam}}\)
\({m_{{O_2}}} = 80,6 - 66,2 = 14,4{\text{ gam}}\)
\( \to {n_{{O_2}}} = \frac{{14,4}}{{32}} = 0,45{\text{ mol}}\)
\( \to V = 0,45.22,4 = 10,08{\text{ lít}}\)
\({n_{{O_2}}} = \frac{1}{2}{n_{KMn{O_4}}} + \frac{3}{2}{n_{KCl{O_3}}}\)
\( \to {n_{KMn{O_4}}} = 0,3{\text{ mol}}\)
\( \to {m_{KMn{O_4}}} = 0,3.158 = 47,4{\text{ gam}}\)
\({m_{KCl{O_3}}} = 0,2.(39 + 35,5 + 16.3) = 24,5{\text{ gam}}\)
\( \to \% {m_{KMn{O_4}}} = 58,8\% ;\% {m_{KCl{O_3}}} = 30,4\% \)
\(\to \% {m_{Mn{O_2}}} = 10,8\%\)
