The volume equivalent of `CO_(2)` (at `STP`) in the reaction,
`NaHCO_(3)+HClrarrNaCl+H_(2)O+CO_(2)` is:

Options

A. `22.4 "litre"`
B. `112 "litre"`
C. `11.2 "litre"`
D. `5.6 "litre"`

Explanation

To solve the question regarding the volume equivalent of \( CO_2 \) at STP in the reaction:

\[ NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2 \]

we will follow these steps:

### Step 1: Identify the Reaction
The reaction shows that sodium bicarbonate (\( NaHCO_3 \)) reacts with hydrochloric acid (\( HCl \)) to produce sodium chloride (\( NaCl \)), water (\( H_2O \)), and carbon dioxide (\( CO_2 \)).

### Step 2: Determine the Moles of \( CO_2 \) Produced
From the balanced chemical equation, we can see that 1 mole of \( NaHCO_3 \) produces 1 mole of \( CO_2 \). Therefore, if we start with 1 mole of \( NaHCO_3 \), we will produce 1 mole of \( CO_2 \).

### Step 3: Use the Volume-Mole Relationship at STP
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 liters. Since we have determined that 1 mole of \( CO_2 \) is produced, we can calculate the volume it occupies.

### Step 4: Calculate the Volume of \( CO_2 \)
Using the relationship:
\[ \text{Volume} = \text{Number of moles} \times 22.4 \text{ L} \]
Substituting the number of moles of \( CO_2 \):
\[ \text{Volume of } CO_2 = 1 \text{ mole} \times 22.4 \text{ L/mole} = 22.4 \text{ L} \]

### Conclusion
Thus, the volume equivalent of \( CO_2 \) produced in the reaction at STP is **22.4 liters**.

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