\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Áp dụng ĐLBT khối lượng:
\(m_{O2\left(thu.duoc\right)}=23,7-22,26=1,44\left(g\right)\)
\(\Rightarrow n_{O2}=\frac{1,44}{32}=0,045\left(mol\right)\)
Theo PTHH,
\(\Rightarrow n_{KMnO4\left(np\right)}=2.0,045=0,09\left(mol\right)\)
\(m_{KMnO4\left(np\right)}=0,09.158=14,22\left(g\right)< 23,7\left(g\right)\)
\(\Rightarrow\) KMnO4 bị nhiệt phân không hoàn toàn
\(m_{KMnO4\left(dư\right)}=23,7-14,22=9,48\left(g\right)\)
\(\Rightarrow n_{K2MnO4}=0,045\left(mol\right)\)
\(\Rightarrow m_{K2MnO4}=0,045.197=8,865\left(g\right)\)
Trong hh X gồm: KMnO4 dư; K2MnO4;MnO2
\(\%_{KMnO4}=\frac{9,48}{22,26}.100\%=42,58\%\)
\(\%_{K2MnO4}=\frac{8,865}{22,26}.100\%=39,83\%\)
\(\%_{MnO2}=100\%-42,58\%-39,83\%=17,59\%\)
\(\Rightarrow H=\frac{14,22}{23,7}.100\%=60\%\)