6. What mass of sodium carbonate must be used to produce 10.36 L of carbon dioxide at $$ 24^{\circ} \mathrm{C} $$ and 103 kPa according to the following neutralization reaction? $$ \mathrm{HCl}_{(\mathrm{aq})}+\mathrm{Na}_{2} \mathrm{CO}_{3(\mathrm{~s})} \rightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} $$

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To solve this problem, we need to determine the mass of sodium carbonate ($$ \mathrm{Na}_2\mathrm{CO}_3 $$) required to produce a specific volume of carbon dioxide ($$ \mathrm{CO}_2 $$) under given conditions of temperature and pressure. We will follow these steps:

1. **Convert the volume of $$ \mathrm{CO}_2 $$ to moles using the Ideal Gas Law.**
2. **Use stoichiometry to find the moles of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ needed.**
3. **Convert moles of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ to mass.**

### Step 1: Calculate Moles of $$ \mathrm{CO}_2 $$

The Ideal Gas Law is given by the equation:
$$
PV = nRT
$$
Where:
- $$ P $$ = pressure in atm
- $$ V $$ = volume in liters
- $$ n $$ = number of moles
- $$ R $$ = ideal gas constant $$ 0.0821 \, \mathrm{L \cdot atm/(K \cdot mol)} $$
- $$ T $$ = temperature in Kelvin

**Convert the given conditions:**
- Pressure: $$ 103 \, \mathrm{kPa} $$ to atm:
$$
P = \frac{103 \, \mathrm{kPa}}{101.325 \, \mathrm{kPa/atm}} \approx 1.016 \, \mathrm{atm}
$$
- Temperature: $$ 24^{\circ} \mathrm{C} $$ to Kelvin:
$$
T = 24 + 273.15 = 297.15 \, \mathrm{K}
$$

**Now, substitute the values into the Ideal Gas Law:**
$$
n = \frac{PV}{RT} = \frac{(1.016 \, \mathrm{atm})(10.36 \, \mathrm{L})}{(0.0821 \, \mathrm{L \cdot atm/(K \cdot mol)})(297.15 \, \mathrm{K})}
$$

### Step 2: Calculate Moles of $$ \mathrm{Na}_2\mathrm{CO}_3 $$

From the balanced chemical equation, we see that 1 mole of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ produces 1 mole of $$ \mathrm{CO}_2 $$. Therefore, the moles of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ required will be equal to the moles of $$ \mathrm{CO}_2 $$.

### Step 3: Convert Moles of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ to Mass

The molar mass of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ is calculated as follows:
- Sodium (Na): $$ 22.99 \, \mathrm{g/mol} \times 2 = 45.98 \, \mathrm{g/mol} $$
- Carbon (C): $$ 12.01 \, \mathrm{g/mol} $$
- Oxygen (O): $$ 16.00 \, \mathrm{g/mol} \times 3 = 48.00 \, \mathrm{g/mol} $$

Total molar mass:
$$
\text{Molar mass of } \mathrm{Na}_2\mathrm{CO}_3 = 45.98 + 12.01 + 48.00 = 105.99 \, \mathrm{g/mol}
$$

Now, we can calculate the mass of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ using the formula:
$$
\text{mass} = n \times \text{molar mass}
$$

Let's perform the calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(1.016\times 10.36\right)}{\left(0.0821\times 297.15\right)}$$
- step1: Remove the parentheses:
  $$\frac{1.016\times 10.36}{0.0821\times 297.15}$$
- step2: Multiply the numbers:
  $$\frac{10.52576}{0.0821\times 297.15}$$
- step3: Multiply the numbers:
  $$\frac{10.52576}{24.396015}$$
- step4: Convert the expressions:
  $$\frac{\frac{32893}{3125}}{\frac{4879203}{200000}}$$
- step5: Multiply by the reciprocal:
  $$\frac{32893}{3125}\times \frac{200000}{4879203}$$
- step6: Reduce the numbers:
  $$4699\times \frac{64}{697029}$$
- step7: Multiply:
  $$\frac{4699\times 64}{697029}$$
- step8: Multiply:
  $$\frac{300736}{697029}$$
Calculate or simplify the expression $$ 0.431454 * 105.99 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$0.431454\times 105.99$$
- step1: Multiply the numbers:
  $$45.72980946$$
### Step 1: Calculate Moles of $$ \mathrm{CO}_2 $$

Using the Ideal Gas Law, we found that the number of moles of $$ \mathrm{CO}_2 $$ produced is approximately:
$$
n \approx 0.4315 \, \mathrm{mol}
$$

### Step 2: Moles of $$ \mathrm{Na}_2\mathrm{CO}_3 $$

Since the stoichiometry of the reaction shows that 1 mole of $$ \mathrm{Na}_2\mathrm{CO}_3 $$ produces 1 mole of $$ \mathrm{CO}_2 $$, we also need:
$$
n(\mathrm{Na}_2\mathrm{CO}_3) \approx 0.4315 \, \mathrm{mol}
$$

### Step 3: Calculate Mass of $$ \mathrm{Na}_2\mathrm{CO}_3 $$

Now, we calculate the mass of $$ \mathrm{Na}_2\mathrm{CO}_3 $$:
$$
\text{mass} = n \times \text{molar mass} = 0.4315 \, \mathrm{mol} \times 105.99 \, \mathrm{g/mol} \approx 45.73 \, \mathrm{g}
$$

### Final Answer

The mass of sodium carbonate required to produce $$ 10.36 \, \mathrm{L} $$ of carbon dioxide at $$ 24^{\circ} \mathrm{C} $$ and $$ 103 \, \mathrm{kPa} $$ is approximately:
$$
\boxed{45.73 \, \mathrm{g}}
$$
To produce 10.36 L of carbon dioxide at 24°C and 103 kPa, approximately 45.73 grams of sodium carbonate are needed.

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