`KO_(2) + CO_(2) + H_(2)O underset(CO_(2))overset("more")(rarr) A + B` , then A and B are respectively

Options

A. `K_(2)CO_(3), O_(2)`
B. `KHCO_(3), O_(2)`
C. `KOH, K_(2)CO_(3)`
D. `KHCO_(3), H_(2)O`

Answer

To solve the given reaction \( KO_2 + CO_2 + H_2O \xrightarrow{CO_2 \text{ (excess)}} A + B \), we will analyze the reactants and determine the products step by step.

### Step 1: Identify the Reactants
The reactants in the equation are:
- \( KO_2 \) (Potassium superoxide)
- \( CO_2 \) (Carbon dioxide)
- \( H_2O \) (Water)

### Step 2: Understand the Reaction Conditions
The reaction occurs in the presence of excess \( CO_2 \). This suggests that \( CO_2 \) will play a significant role in the products formed.

### Step 3: Analyze the Reaction
When potassium superoxide reacts with carbon dioxide and water, it typically produces potassium hydrogen carbonate (also known as potassium bicarbonate) and oxygen gas. The reaction can be summarized as follows:

\[ 2 KO_2 + 2 CO_2 + 2 H_2O \rightarrow 2 KHC(O_3) + O_2 \]

### Step 4: Identify the Products
From the reaction:
- **Product A** is potassium hydrogen carbonate, which can be represented as \( KHC(O_3) \) or \( KHCO_3 \).
- **Product B** is oxygen gas, represented as \( O_2 \).

### Conclusion
Thus, the products \( A \) and \( B \) are:
- \( A = KHCO_3 \) (Potassium hydrogen carbonate)
- \( B = O_2 \) (Oxygen gas)

### Final Answer
Therefore, the final products are:
- A = \( KHCO_3 \)
- B = \( O_2 \)

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