Ta có pthh 1
2KMnO4 + 16HCl \(\rightarrow\) 2KCl + 2MnCl2 + 8H2O +5Cl2
Pthh 2
3Cl2 + 2Fe \(\rightarrow\) 2FeCl3
Theo đề bài ta có
nFeCl3=\(\dfrac{16,25}{162,5}=0,1mol\)
Theo pthh 2
nCl2=\(\dfrac{3}{2}nFeCl3=\dfrac{3}{2}.0,1=0,15mol\)
Theo pthh 1
nKMnO4=\(\dfrac{2}{5}nCl2=\dfrac{2}{5}.0,15=0,06mol\)
\(\Rightarrow\) mKMnO4=0,06.158=9,48 g