Percentage loss in mass, when `NaHCO_(3)(s)` is heated in open vessel `2NaHCO_(3)(s)rarrNa_(2)CO_(3)(s)+CO_(2)(g)+H_(2)O(g)`

Options

A. `21.12%`
B. `36.9%`
C. `30%`
D. `32.23%`

Answer

To find the percentage loss in mass when sodium bicarbonate (NaHCO₃) is heated in an open vessel, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction for the thermal decomposition of sodium bicarbonate is:
\[ 2 \text{NaHCO}_3 (s) \rightarrow \text{Na}_2\text{CO}_3 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (g) \]

### Step 2: Determine the molar masses
- Molar mass of NaHCO₃:
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (4 O in total)

\[ \text{Molar mass of NaHCO}_3 = 23 + 1 + 12 + (3 \times 16) = 84 \text{ g/mol} \]

- Molar mass of CO₂:
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (2 O in total)

\[ \text{Molar mass of CO}_2 = 12 + (2 \times 16) = 44 \text{ g/mol} \]

- Molar mass of H₂O:
- Hydrogen (H): 1 g/mol (2 H in total)
- Oxygen (O): 16 g/mol

\[ \text{Molar mass of H}_2\text{O} = (2 \times 1) + 16 = 18 \text{ g/mol} \]

### Step 3: Calculate the mass loss
From the balanced equation, 2 moles of NaHCO₃ produce:
- 1 mole of CO₂
- 1 mole of H₂O

For 1 mole of NaHCO₃:
- CO₂ produced = 0.5 moles
- H₂O produced = 0.5 moles

Now, we can calculate the mass loss:
- Mass loss due to CO₂:
\[ \text{Mass of CO}_2 = 0.5 \times 44 \text{ g} = 22 \text{ g} \]
- Mass loss due to H₂O:
\[ \text{Mass of H}_2\text{O} = 0.5 \times 18 \text{ g} = 9 \text{ g} \]

Total mass loss:
\[ \text{Total mass loss} = 22 \text{ g} + 9 \text{ g} = 31 \text{ g} \]

### Step 4: Calculate the percentage loss in mass
To find the percentage loss in mass:
\[ \text{Percentage loss} = \left( \frac{\text{Mass loss}}{\text{Initial mass}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage loss} = \left( \frac{31 \text{ g}}{84 \text{ g}} \right) \times 100 \approx 36.9\% \]

### Final Answer
The percentage loss in mass when NaHCO₃ is heated in an open vessel is approximately **36.9%**.

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