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 find the mass of \( CO_2 \) released in the reaction between sodium bicarbonate (\( NaHCO_3 \)) and acetic acid (\( CH_3COOH \)), we can follow these steps:

### Step 1: Calculate the total initial mass
First, we need to calculate the total mass of the reactants before the reaction occurs. This includes the mass of sodium bicarbonate and the mass of acetic acid solution.

\[ \text{Total mass of reactants} = \text{mass of } NaHCO_3 + \text{mass of } CH_3COOH \]
\[ \text{Total mass of reactants} = 12.6 \, \text{g} + 30.0 \, \text{g} = 42.6 \, \text{g} \]

### Step 2: Set up the equation based on the law of conservation of mass
According to the law of conservation of mass, the total mass before the reaction must equal the total mass after the reaction. The mass after the reaction consists of the residue and the mass of \( CO_2 \) released.

Let \( W \) be the mass of \( CO_2 \) released. Therefore, we can write the equation as:

\[ \text{Total mass after reaction} = \text{mass of residue} + W \]
\[ 42.6 \, \text{g} = 36.0 \, \text{g} + W \]

### Step 3: Solve for the mass of \( CO_2 \)
Now, we can rearrange the equation to solve for \( W \):

\[ W = 42.6 \, \text{g} - 36.0 \, \text{g} \]
\[ W = 6.6 \, \text{g} \]

### Conclusion
The mass of \( CO_2 \) released in the reaction is \( 6.6 \, \text{g} \).

---