If 12.6 g of `NaHCO_(3)` are added to 30.0 g of `CH_(3)COOH` solution, the residue is found is found to weight 36.0 g. What is the mass `CO_(2)` released in the reaction ?

Answer

To solve the problem, we need to determine the mass of carbon dioxide (CO₂) released when sodium bicarbonate (NaHCO₃) reacts with acetic acid (CH₃COOH).

### Step-by-step Solution:

1. **Identify the reaction**: The reaction between sodium bicarbonate and acetic acid can be represented as:
\[ \text{NaHCO}_3 + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \]
This shows that sodium bicarbonate reacts with acetic acid to produce sodium acetate, water, and carbon dioxide.

2. **Write down the given data**:
- Mass of NaHCO₃ = 12.6 g
- Mass of CH₃COOH = 30.0 g
- Mass of residue (products) = 36.0 g

3. **Use the law of conservation of mass**: According to this law, the total mass of reactants equals the total mass of products. We can express this as:
\[ \text{Mass of reactants} = \text{Mass of products} + \text{Mass of CO}_2 \]
Let the mass of CO₂ released be \( x \) grams. Therefore, we can write:
\[ 12.6 \, \text{g} + 30.0 \, \text{g} = 36.0 \, \text{g} + x \]

4. **Calculate the total mass of reactants**:
\[ 12.6 \, \text{g} + 30.0 \, \text{g} = 42.6 \, \text{g} \]

5. **Set up the equation**:
\[ 42.6 \, \text{g} = 36.0 \, \text{g} + x \]

6. **Solve for \( x \)**:
\[ x = 42.6 \, \text{g} - 36.0 \, \text{g} = 6.6 \, \text{g} \]

7. **Conclusion**: The mass of carbon dioxide released in the reaction is:
\[ \boxed{6.6 \, \text{g}} \]