a)\(2KMnO4-->K2MnO4+MnO2+O2\)
\(n_{KMnO4}=\frac{15,8}{158}=0,1\left(mol\right)\)
\(n_{O2}=\frac{1}{2}n_{KMnO4}=0,05\left(mol\right)\)
\(V=V_{O2}=0,05.22,4=1,12\left(l\right)\)
b) Lấy 1/5 V O2 => \(n=\frac{1}{5}n_{O2}=0,01\left(mol\right)\)
\(n_{Na}=\frac{1,38}{23}=0,06\left(mol\right)\)
\(4Na+O2--.2Na2O\)
Lập tỉ lệ
\(n_{Na}\left(\frac{0,06}{4}\right)>n_{O2}\left(\frac{0,01}{1}\right)=>Nadư\)
\(n_{Na}=4n_{O2}=0,04\left(mol\right)\)
\(\Rightarrow n_{Na}dư=0,06-0,04=0,02\left(mol\right)\)
\(m_{Na}dư=0,02.23=0,46\left(g\right)\)
