Which of the following `is//are` correct?
The following reaction occurs:
`Na_(2) CO_(3) + 2HCl rarr 2NACl + CO_(2) + H_(2) O`
`106.g "of" Na_(2) CO_(3)` reacts with `109.5 g "of" HCl`.

Options

A. The `HCl` is in excess.
B. `117.0 g` of `NaCl` is formed.
C. The volume of `CO_(2)` produced at 1 bar and `273 K` is `22.7 L`
D. The volume of `CO_(2)` produced at 1 bar and `298 K` is `24.7 L`

Explanation

To solve the problem, we need to analyze the given chemical reaction and perform calculations based on the provided masses of reactants. The reaction is:

\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 1: Calculate the Molar Mass of Reactants
1. **Molar Mass of Na2CO3**:
- Sodium (Na): 23 g/mol, Carbon (C): 12 g/mol, Oxygen (O): 16 g/mol
- Molar mass of Na2CO3 = \( 2 \times 23 + 12 + 3 \times 16 = 46 + 12 + 48 = 106 \, \text{g/mol} \)

2. **Molar Mass of HCl**:
- Hydrogen (H): 1 g/mol, Chlorine (Cl): 35.5 g/mol
- Molar mass of HCl = \( 1 + 35.5 = 36.5 \, \text{g/mol} \)

### Step 2: Calculate Moles of Reactants
1. **Moles of Na2CO3**:
- Given mass = 106 g
- Moles of Na2CO3 = \( \frac{106 \, \text{g}}{106 \, \text{g/mol}} = 1 \, \text{mol} \)

2. **Moles of HCl**:
- Given mass = 109.5 g
- Moles of HCl = \( \frac{109.5 \, \text{g}}{36.5 \, \text{g/mol}} \approx 3 \, \text{mol} \)

### Step 3: Determine Limiting Reactant
From the balanced equation, we see that:
- 1 mole of Na2CO3 reacts with 2 moles of HCl.

For 1 mole of Na2CO3, we need:
- 2 moles of HCl.

Since we have 1 mole of Na2CO3 and 3 moles of HCl:
- HCl is in excess (3 moles available, but only 2 moles are required).
- Therefore, Na2CO3 is the limiting reactant.

### Step 4: Calculate Amount of Products Formed
1. **Moles of NaCl Formed**:
- From the reaction, 1 mole of Na2CO3 produces 2 moles of NaCl.
- Therefore, moles of NaCl = \( 1 \times 2 = 2 \, \text{mol} \)

2. **Mass of NaCl**:
- Molar mass of NaCl = \( 23 + 35.5 = 58.5 \, \text{g/mol} \)
- Mass of NaCl = \( 2 \, \text{mol} \times 58.5 \, \text{g/mol} = 117 \, \text{g} \)

3. **Moles of CO2 Formed**:
- From the reaction, 1 mole of Na2CO3 produces 1 mole of CO2.
- Therefore, moles of CO2 = 1 mole.

### Step 5: Calculate Volume of CO2 at STP
1. **Volume of CO2**:
- At STP, 1 mole of any gas occupies 22.4 liters.
- Volume of CO2 = \( 1 \, \text{mol} \times 22.4 \, \text{L/mol} = 22.4 \, \text{L} \)

### Summary of Results
- Moles of Na2CO3 = 1 mol
- Moles of HCl = 3 mol
- Limiting reactant = Na2CO3
- Mass of NaCl produced = 117 g
- Volume of CO2 produced = 22.4 L

### Conclusion
All options provided in the question are correct based on the calculations performed.

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