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

Options

A. 15.78 mL
B. 157.8 mL
C. 198.4 mL
D. 195.5 mL

Explanation

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

### Step 1: Define the weights of the components in the mixture
Let the weight of Na₂CO₃ be \( x \) grams. Then, the weight of NaHCO₃ will be \( 1 - x \) grams since the total weight of the mixture is 1 gram.

### Step 2: Calculate the molar masses
- Molar mass of Na₂CO₃ (Sodium Carbonate) = 106 g/mol
- Molar mass of NaHCO₃ (Sodium Bicarbonate) = 84 g/mol

### Step 3: Set up the equations for moles
The number of moles of Na₂CO₃:
\[ \text{Moles of Na₂CO₃} = \frac{x}{106} \]

The number of moles of NaHCO₃:
\[ \text{Moles of NaHCO₃} = \frac{1 - x}{84} \]

### Step 4: Use the equimolecular condition
Since the mixture is equimolecular, we set the number of moles equal:
\[ \frac{x}{106} = \frac{1 - x}{84} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ 84x = 106(1 - x) \]
\[ 84x = 106 - 106x \]
\[ 84x + 106x = 106 \]
\[ 190x = 106 \]
\[ x = \frac{106}{190} \approx 0.558 \text{ grams} \]

### Step 6: Calculate the weight of NaHCO₃
\[ \text{Weight of NaHCO₃} = 1 - x = 1 - 0.558 = 0.442 \text{ grams} \]

### Step 7: Calculate the number of moles of each compound
- Moles of Na₂CO₃:
\[ \text{Moles of Na₂CO₃} = \frac{0.558}{106} \approx 0.00526 \text{ moles} \]

- Moles of NaHCO₃:
\[ \text{Moles of NaHCO₃} = \frac{0.442}{84} \approx 0.00526 \text{ moles} \]

### Step 8: Determine the moles of HCl required
1. For Na₂CO₃:
- The reaction is:
\[ \text{Na₂CO₃} + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H₂O} + \text{CO₂} \]
- Moles of HCl required:
\[ 2 \times 0.00526 = 0.01052 \text{ moles} \]

2. For NaHCO₃:
- The reaction is:
\[ \text{NaHCO₃} + \text{HCl} \rightarrow \text{NaCl} + \text{H₂O} + \text{CO₂} \]
- Moles of HCl required:
\[ 0.00526 \text{ moles} \]

### Step 9: Total moles of HCl required
\[ \text{Total moles of HCl} = 0.01052 + 0.00526 = 0.01578 \text{ moles} \]

### Step 10: Calculate the volume of HCl solution needed
Using the molarity formula:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
We rearrange to find the volume:
\[ \text{Volume} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.01578}{0.1} = 0.1578 \text{ liters} = 157.8 \text{ mL} \]

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

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