19 g of a mixture containing `NaHCO_(3)` and `Na_(2)CO_(3)` on complete heating liberated 1, 12 L of `CO_(2)` at STP. The weight (in g) of `Na_(2)CO_(3)` in the mixture before heating ?

Options

A. 8.4
B. 15.9
C. `4.0`
D. `10.6`

Explanation

To solve the problem, we need to find the weight of `Na2CO3` in a mixture that initially contained `NaHCO3` and `Na2CO3`, given that the mixture released `1.12 L` of `CO2` upon heating.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The decomposition reaction for sodium bicarbonate (`NaHCO3`) upon heating is:
\[ 2 \, \text{NaHCO}_3 \, (s) \rightarrow \text{Na}_2\text{CO}_3 \, (s) + \text{H}_2\text{O} \, (g) + \text{CO}_2 \, (g) \]
This means that 2 moles of `NaHCO3` produce 1 mole of `Na2CO3` and 1 mole of `CO2`.

2. **Calculate Moles of CO2 Produced**:
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies `22.4 L`. Therefore, we can calculate the moles of `CO2` produced:
\[ \text{Moles of } CO_2 = \frac{\text{Volume of } CO_2}{22.4 \, L} = \frac{1.12 \, L}{22.4 \, L/mol} = 0.05 \, mol \]

3. **Relate Moles of NaHCO3 to Moles of CO2**:
From the balanced equation, 1 mole of `CO2` is produced from 2 moles of `NaHCO3`. Therefore, the moles of `NaHCO3` that decomposed can be calculated as:
\[ \text{Moles of } NaHCO_3 = 2 \times \text{Moles of } CO_2 = 2 \times 0.05 \, mol = 0.1 \, mol \]

4. **Calculate Mass of NaHCO3**:
The molar mass of `NaHCO3` is calculated as follows:
- Na: 23 g/mol
- H: 1 g/mol
- C: 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of } NaHCO_3 = 23 + 1 + 12 + 48 = 84 \, g/mol \]
Now, we can find the mass of `NaHCO3` that decomposed:
\[ \text{Mass of } NaHCO_3 = \text{Moles} \times \text{Molar Mass} = 0.1 \, mol \times 84 \, g/mol = 8.4 \, g \]

5. **Calculate Mass of Na2CO3 in the Mixture**:
The total mass of the mixture is given as `19 g`. Therefore, the mass of `Na2CO3` in the mixture can be calculated as:
\[ \text{Mass of } Na_2CO_3 = \text{Total mass} - \text{Mass of } NaHCO_3 = 19 \, g - 8.4 \, g = 10.6 \, g \]

### Final Answer:
The weight of `Na2CO3` in the mixture before heating is **10.6 g**.