JEE MAIN - Chemistry (2019 - 12th April Evening Slot - No. 14)
Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and gaseous product.
MnO2 reacts with NaCl and concentrated H2O4 to give a pungent gas Z. X, Y and Z, respectively, are :
KMnO4, K2MnO4 and Cl2
K2MnO4, KMnO4 and SO2
K3MnO4, K2MnO4 and Cl2
K2MnO4, KMnO4 and Cl2
Explanation
| KMnO4 | $$\buildrel {513\,\,K} \over \longrightarrow $$ | K2MnO4 | + | MnO2 | + | O2 |
|---|---|---|---|---|---|---|
| (X) | (Y) |
| MnO2 | + | NaCl | + | conc H2SO4 | $$ \to $$ | MnSO4 | + | NaHSO4 | + | H2O | + | Cl2 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| (Z) |
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