A compound of sodium does not give `CO_(2)` when heated but it gives `CO_(2)` when treated with dilute acids. A crystalline compound is found to have 37.1% Na and 14.52% `H_(2)O`. Hence, compound is

Options

A. `NaHCO_(3)*10H_(2)O`
B. `NaHCO_(3)*5H_(2)O`
C. `Na_(2)CO_(3)*10H_(2)O`
D. `Na_(2)CO_(3)*H_(2)O`

Explanation

To solve the problem step by step, we will analyze the information provided and calculate the necessary values to identify the compound.

### Step 1: Analyze the Information Given
The compound of sodium does not give CO2 when heated but does give CO2 when treated with dilute acids. This suggests that the compound is likely sodium carbonate (Na2CO3) because it reacts with acids to produce CO2.

### Step 2: Determine the Composition of the Compound
We are given that the compound has:
- 37.1% Sodium (Na)
- 14.52% Water (H2O)

### Step 3: Calculate the Percentage of Na2CO3
Since the total percentage of the compound must equal 100%, we can find the percentage of Na2CO3:
- Percentage of Na2CO3 = 100% - Percentage of H2O
- Percentage of Na2CO3 = 100% - 14.52% = 85.48%

### Step 4: Set Up the Molar Ratios
Let’s denote the compound as Na2CO3·xH2O. We need to find the molar ratio of Na2CO3 to H2O.

1. Calculate the molar mass of Na2CO3:
- Na: 23 g/mol × 2 = 46 g/mol
- C: 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 46 + 12 + 48 = 106 g/mol

2. Calculate the molar mass of H2O:
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol
- Total = 2 + 16 = 18 g/mol

### Step 5: Calculate the Mass of Each Component
Assuming we have 100 g of the compound:
- Mass of Na2CO3 = 85.48 g
- Mass of H2O = 14.52 g

### Step 6: Calculate Moles of Each Component
1. Moles of Na2CO3:
- Moles = mass/molar mass = 85.48 g / 106 g/mol ≈ 0.806 moles

2. Moles of H2O:
- Moles = mass/molar mass = 14.52 g / 18 g/mol ≈ 0.807 moles

### Step 7: Find the Simplified Ratio
Now we can find the ratio of moles of Na2CO3 to moles of H2O:
- Ratio = 0.806 : 0.807 ≈ 1 : 1

### Step 8: Determine the Hydration State
Since the ratio of Na2CO3 to H2O is approximately 1:1, we can conclude that the compound is Na2CO3·H2O.

### Step 9: Verify the Percentage of Sodium
To confirm, we calculate the percentage of sodium in Na2CO3·H2O:
- Total molar mass of Na2CO3·H2O = 106 g/mol + 18 g/mol = 124 g/mol
- Percentage of Na = (46 g / 124 g) × 100% ≈ 37.1%

### Conclusion
The compound is sodium carbonate monohydrate, or Na2CO3·H2O.

---