\(V_{O2_{thu.duoc}}=168.10=1680\left(ml\right)=1,68\left(l\right)\)
\(\Rightarrow V_{O2_{tao.ra}}=\frac{1,68}{75\%}=2,24\left(l\right)\)
\(\Rightarrow n_{O2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
\(\Rightarrow n_{KMnO4}=2n_{O2}=0,1.2=0,2\left(mol\right)\)
\(\Rightarrow m_{KMnO4}=0,2.\left(39+55+16.4\right)=31,6\left(g\right)\)