Đáp án:
\({m_{KMn{O_4}}} = 189,6{\text{ gam}}\)
\({m_{{K_2}Mn{O_4}}} = 118,2{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
Ta có:
\({n_{{O_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol}}\)
\( \to {n_{{K_2}Mn{O_4}}} = {n_{{O_2}}} = 0,6{\text{ mol}}\)
\({n_{KMn{O_4}}} = 2{n_{{O_2}}} = 1,2{\text{ mol}}\)
\( \to {m_{KMn{O_4}}} = 1,2.(39 + 55 + 16.4) = 189,6{\text{ gam}}\)
\({m_{{K_2}Mn{O_4}}} = 0,6.(39.2 + 55 + 16.4) = 118,2{\text{ gam}}\)
