Đáp án:
a) KMnO4: 30,76%; KClO3: 69,24%
b) 124,608 g
Giải thích các bước giải:
a) $\begin{gathered}
{m_{Mn}} = \dfrac{{308,2.10,69}}{{100}} = 32,95g \Rightarrow {n_{Mn}} = 0,6mol \hfill \\
\Rightarrow {n_{KMn{O_4}}} = {n_{Mn}} = 0,6mol \hfill \\
\Rightarrow \% {m_{KMn{O_4}}} = \dfrac{{0,6.158}}{{308,2}}.100\% = 30,76\% \hfill \\
\Rightarrow \% {m_{KCl{O_3}}} = 100 - 30,76 = 69,24\% \hfill \\
\end{gathered} $
b)
$\begin{gathered}
{m_{KCl{O_3}}} = 308,2 - 0,6.158 = 213,4g \hfill \\
\Rightarrow {n_{KCl{O_3}}} = \dfrac{{213,4}}{{122,5}} = 1,742mol \hfill \\
\end{gathered} $
$\begin{gathered}
2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2} \hfill \\
2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2} \hfill \\
\end{gathered} $
$\sum {{n_{{O_2}}} = \dfrac{1}{2}{n_{KMn{O_4}}} + \dfrac{3}{2}{n_{KCl{O_3}}} = \dfrac{1}{2}.0,6 + \dfrac{3}{2}.1,742 = 2,913mol} $
${n_{Mg}} = \dfrac{{78}}{{24}} = 3,25mol;{n_{{O_2}}} = \dfrac{{2,913}}{2} = 1,4565mol$
$2Mg + {O_2} \to 2MgO$
$\dfrac{{{n_{Mg}}}}{2} = 1,625 > \dfrac{{{n_{{O_2}}}}}{1} = 1,4565$ ⇒ $Mg$ dư, ${O_2}$ hết
$\begin{gathered}
{n_{Mgpu}} = {n_{MgO}} = 2{n_{{O_2}}} = 2.1,4565 = 2,913mol \hfill \\
{n_{Mgdu}} = 3,25 - 2,913 = 0,337mol \hfill \\
\end{gathered} $
$ \Rightarrow m = {m_{MgO}} + {m_{Mgdu}} = 2,913.40 + 0,337.24 = 124,608g$
