MI=4,4375.16=71(Cl2)
nL=nHCl=0,2.0,25=0,05mol
Do L là bazo kiem LOH nên MLOH=\(\dfrac{2,8}{0,05}=56\rightarrow KOH\)
2KMnO4(A)\(\overset{t^0}{\rightarrow}\)K2MnO4(B)+MnO2(C)+O2(D)
MnO2(C)+4HCl(E)\(\rightarrow\)MnCl2(G)+2H2O(H)+Cl2(I)
2KMnO4(A)+16HCl(E)\(\rightarrow\)2KCl(K)+2MnCl2(G)+5Cl2(I)+8H2O(H)
2KCl(K)+2H2O(H)\(\rightarrow\)2KOH(L)+Cl2(I)+H2(M)