One gram equimolecular mixture of `Na_(2)CO_(3)` and `NaHCO_(3)`is reacted with 0.1 NHCl. The milliliters of 0.1 N HCl required to react completely with the above mixture is :

Options

A. 15.78 mL
B. 157.8 mL
C. 198.4 mL
D. 308 mL

Answer

To solve the problem of how many milliliters of 0.1 N HCl are required to react completely with a 1 gram equimolecular mixture of Na₂CO₃ and NaHCO₃, we can follow these steps:

### Step 1: Determine the Molar Masses
First, we need to know the molar masses of the compounds involved:
- Molar mass of Na₂CO₃ (Sodium Carbonate) = 2(23) + 12 + 3(16) = 106 g/mol
- Molar mass of NaHCO₃ (Sodium Bicarbonate) = 23 + 1 + 12 + 3(16) = 84 g/mol

### Step 2: Set Up the Equimolecular Mixture
Since we have an equimolecular mixture, let the number of moles of Na₂CO₃ be \( x \) and the number of moles of NaHCO₃ also be \( x \).

### Step 3: Write the Mass Equation
The total mass of the mixture is given as 1 gram:
\[ x \cdot 106 + x \cdot 84 = 1 \]
\[ x(106 + 84) = 1 \]
\[ x(190) = 1 \]
\[ x = \frac{1}{190} \text{ moles} \]

### Step 4: Calculate the Number of Moles
Now, calculate the number of moles:
\[ x = \frac{1}{190} \approx 0.00526 \text{ moles} \]

### Step 5: Determine the Equivalent Moles
Next, we find the equivalent moles of each compound:
- For Na₂CO₃, the n-factor is 2 (it can donate 2 Na⁺ ions).
- For NaHCO₃, the n-factor is 1 (it can donate 1 Na⁺ ion).

Thus, the equivalents contributed by each:
- Equivalents from Na₂CO₃ = \( 2x = 2 \times 0.00526 = 0.01052 \) equivalents
- Equivalents from NaHCO₃ = \( 1x = 0.00526 \) equivalents

Total equivalents = \( 0.01052 + 0.00526 = 0.01578 \) equivalents.

### Step 6: Relate to HCl
Since HCl is a strong acid that provides 1 equivalent of H⁺ per mole, the total equivalents of HCl required will be equal to the total equivalents from the mixture:
\[ \text{Total equivalents of HCl} = 0.01578 \]

### Step 7: Calculate Volume of HCl
Using the normality equation:
\[ N \cdot V = \text{equivalents} \]
Where \( N = 0.1 \) N (normality of HCl):
\[ 0.1 \cdot V = 0.01578 \]
\[ V = \frac{0.01578}{0.1} = 0.1578 \text{ L} = 157.8 \text{ mL} \]

### Final Answer
The volume of 0.1 N HCl required to react completely with the mixture is **157.8 mL**.

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