Q. 71

A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

Answer :

A=KMnO4, B=K2MnO4, C= MnO2, D=MnCl2

Explanation:


A violet compound of manganese which is potassium permanganate (KMnO4), decomposes to liberate potassium manganate(K2MnO4) and manganese dioxide(MnO2) along with oxygen.


KMnO4(A) K2MnO4(B) +MnO2(C) + O2


Manganese dioxide (MnO2)reacts with KOH to give potassium manganate (K2MnO4)


MnO2(C) + KOH + O22K2MnO4(B) +2H2O


On heating Manganese dioxide (MnO2) with NaCl and H2SO4, we get Manganese(II) chloride(MnCl2), chlorine gas and other products,


MnO2(C) + 4NaCl + 4H2SO4 MnCl2(D) +4NaHSO4+2H2O +Cl2


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