Đáp án:
\(\% {n_{KCl{O_3}}} = 61,5\%; \% {n_{KMn{O_4}}} = 38,5\% \)
\(\% {m_{KCl}} = 45,63\% ;\% {m_{{K_2}Mn{O_4}}} = 37,71\% ;\% {m_{Mn{O_2}}} = 16,66\% \)
Giải thích các bước giải:
Gọi số mol \(KClO_3;KMnO_4\) lần lượt là \(x;y\)
\( \to 122,5x + 158y = 53,1\)
Phản ứng xảy ra:
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
Ta có:
\({n_{{O_2}}} = \frac{3}{2}{n_{KCl{O_3}}} + \frac{1}{2}{n_{KMn{O_4}}} = 1,5x + 0,5y = \frac{{9,744}}{{22,4}} = 0,435{\text{ mol}}\)
Giải được:
\(x=0,24;y=0,15\)
\( \to \% {n_{KCl{O_3}}} = \frac{{0,24}}{{0,24 + 0,15}} = 61,5\% \to \% {n_{KMn{O_4}}} = 38,5\% \)
Ta có:
\({n_{KCl}} = {n_{KCl{O_3}}} = 0,24{\text{ mol;}}{{\text{n}}_{{K_2}Mn{O_4}}} = {n_{Mn{O_2}}} = \frac{1}{2}{n_{KMn{O_4}}} = 0,075{\text{ mol}}\)
\( \to {m_{KCl}} = 0,24.(39 + 35,5) = 17,88{\text{ gam}}\)
\({m_{{K_2}Mn{O_4}}} = 0,075.(39.2 + 55 + 16.4) = 14,775{\text{ gam}}\)
\( \to {m_{Mn{O_2}}} = 0,075.(55 + 16.2) = 6,525{\text{ gam}}\)
\( \to \% {m_{KCl}} = 45,63\% ;\% {m_{{K_2}Mn{O_4}}} = 37,71\% ;\% {m_{Mn{O_2}}} = 16,66\% \)
