Thermal decomposition of a Mn compound (X) at 513K results in...

Potassium manganate is water soluble and so is leached out as green filtrate. MnO₂ is insoluble in water and gives black residue. The oxygen gas is released in this reaction.

In the first step potassium permanganate $( {KMnO}_{4} )$ the compound (X) decomposes on heating to give potassium manganate $( {K}_{2}{MnO}_{4} )$ which is the compound (Y), ${MnO}_{2}$and the gaseous product is oxygen.

In the second step ${MnO}_{2}$ further reacts with $NaCl$ and concentrated sulfuric acid to give a pungent-smelling gas that is Chlorine gas $( {Cl}_{2} ) , {MnSO}_{4}, {Na}_{2}{SO}_{4}$ and ${H}_{2}O$.

$\mathop{2{KMnO}_{4}}_{\left(X\right)}\mathop{\to}_{\Delta}^{513K}\Delta{K}_{2}\mathop{Mn}_{( Y )}{O}_{4} + {MnO}_{2} + {O}_{2\left(g\right)}$

${MnO}_{2} + 4NaCl + 4{H}_{2}{SO}_{4} \to {MnCl}_{2} + 4{NaHSO}_{4} + 2{H}_{2}O + \mathop{{Cl}_{2\left(g\right)}}_{\left(Z\right)Pungentgas}$

The compound$X = {KMnO}_{4}, Y = {K}_{2}{MnO}_{4}, Z = {Cl}_{2}$.