Đáp án đúng:
Giải chi tiết:\(\eqalign{ & \left\{ \matrix{ KMn{O_4} \hfill \cr KCl{O_3} \hfill \cr} \right.\buildrel {{t^o}} \over \longrightarrow \left\{ \matrix{ {O_2} + kk\left\{ \matrix{ {O_2} \hfill \cr {N_2} \hfill \cr} \right. \to C\left\{ \matrix{ {O_2} \hfill \cr {N_2} \hfill \cr} \right.\buildrel { + C:\,0,044} \over \longrightarrow \uparrow \left\{ \matrix{ {O_2}\,du \hfill \cr C{O_2} \hfill \cr {N_2} \hfill \cr} \right. = > \% O2\,du = 17,083\% \hfill \cr ran\,B(KMn{O_4},{K_2}Mn{O_4},Mn{O_2},\mathop {KCl}\limits_{0,012} ) \hfill \cr} \right. \cr & G/S:\,\underbrace {{O_2}}_{a\,mol}\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{1:3}^{{O_2}\,kk}} kk\left\{ \matrix{ {O_2}:0,6a \hfill \cr {N_2}:2,4a \hfill \cr} \right. \to C\left\{ \matrix{ {O_2}:1,6a \hfill \cr {N_2}:2,4a \hfill \cr} \right. \to \underbrace {\left\{ \matrix{ \buildrel {BTNT\,\,C} \over \longrightarrow C{O_2}:0,044 \hfill \cr \buildrel {BTNT\,\,O} \over \longrightarrow {O_2}\,du:1,6a - 0,044 \hfill \cr {N_2}:2,4a \hfill \cr} \right.}_{4a(mol)} \cr & \to {\rm{ }}1,6a{\rm{ }}-{\rm{ }}0,044{\rm{ }} = {\rm{ }}17,083\% .4a{\rm{ }} \to {\rm{ }}a{\rm{ }} = {\rm{ }}0,048 \cr & \left. \matrix{ KCl\buildrel {BTNT\,Cl} \over \longrightarrow KCl{O_3} \hfill \cr BTKL:{m_A} = {m_{{O_2}}} + {m_B} = 12,536 \hfill \cr} \right\} \to KMn{O_4} \to \left\{ \matrix{ \% KMn{O_4} = 88,226\% \hfill \cr \% KCl{O_3}:11,774\% \hfill \cr} \right. \cr & B\left\{ \matrix{ KMn{O_4},{K_2}Mn{O_4},Mn{O_2} \hfill \cr KCl:1,012 \hfill \cr} \right.\buildrel { + {H_2}S{O_4}} \over \longrightarrow \cr} \)