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Oxalate +MnO(2)+"dil."H(2)SO(4)rarr Gas...

Oxalate `+MnO_(2)+"dil."H_(2)SO_(4)rarr ` Gas. The gas evolved is

A

`CO_(2)`

B

`CO`

C

`SO_(2)`

D

`O_(2)`

Text Solution

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The correct Answer is:
To determine the gas evolved when oxalate reacts with manganese dioxide in the presence of dilute sulfuric acid, we can follow these steps: ### Step 1: Identify the Reactants The reactants in the reaction are: - Oxalate ion (C2O4^2-) - Manganese dioxide (MnO2) - Dilute sulfuric acid (H2SO4) ### Step 2: Write the Reaction When oxalate reacts with manganese dioxide in dilute sulfuric acid, the reaction can be represented as: \[ \text{C}_2\text{O}_4^{2-} + \text{MnO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{MnSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \] ### Step 3: Identify the Products From the reaction, we can identify the products formed: - Manganese sulfate (MnSO4) - Carbon dioxide (CO2) - Water (H2O) ### Step 4: Identify the Evolved Gas The gas evolved in this reaction is carbon dioxide (CO2). This is evident from the products formed during the reaction. ### Conclusion Thus, the gas evolved when oxalate reacts with manganese dioxide in the presence of dilute sulfuric acid is **carbon dioxide (CO2)**. ---
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