2KMnO4 \(\underrightarrow{to}\) K2MnO4 + MnO2 + O2
\(n_{KMnO_4}=\frac{63,2}{158}=0,4\left(mol\right)\)
a) \(m_{O_2}=63,2-57,44=5,76\left(g\right)\)
\(\Rightarrow n_{O_2}=\frac{5,76}{32}=0,18\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,18\times22,4=4,032\left(l\right)\)
b) A gồm: KMnO4 dư, K2MnO4, MnO2
Theo pT: \(n_{KMnO_4}pư=2n_{O_2}=2\times0,18=0,36\left(mol\right)\)
\(\Rightarrow n_{KMnO_4}dư=0,4-0,36=0,04\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,04\times158=6,32\left(g\right)\)
Theo pT: \(n_{K_2MnO_4}=n_{O_2}=0,18\left(mol\right)\)
\(\Rightarrow m_{K_2MnO_4}=0,18\times197=35,46\left(g\right)\)
Theo PT: \(n_{MnO_2}=n_{O_2}=0,18\left(mol\right)\)
\(\Rightarrow m_{MnO_2}=0,18\times87=15,66\left(g\right)\)
Ta có: \(m_A=6,32+35,46+15,66=57,44\left(g\right)\)
\(\%KMnO_4dư=\frac{6,32}{57,44}\times100\%=11\%\)
\(\%K_2MnO_4=\frac{35,46}{57,44}\times100\%=61,73\%\)
\(\%MnO_2=100\%-11\%-61,73\%=27,27\%\)
