Question

4.0 g of a mixture containing Na₂CO₃ and NaHCO₃ is heated to 673K. Loss in mass of the mixture is found to be 0.62g. The percentage of sodium carbonate in the mixture is

• A. 42
• B. 58
• C. 48
• D. 52

Hint:

In problems involving heating of mixtures of carbonates and bicarbonates of alkali metals, remember that bicarbonates decompose easily, while carbonates (except Li₂CO₃) are thermally stable. The mass loss is always due to the decomposition of the bicarbonate component.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a problem based on stoichiometry and thermal decomposition. When the mixture of sodium carbonate (Na₂CO₃) and sodium bicarbonate (NaHCO₃) is heated, only sodium bicarbonate decomposes. Sodium carbonate is thermally stable at 673K and does not decompose. The loss in mass is due to the gaseous products (CO₂ and H₂O) escaping.
Step 2: Key Formula or Approach:
The balanced chemical equation for the decomposition of NaHCO₃ is:
\[ 2 \text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g) \] We will use this equation to relate the mass of NaHCO₃ decomposed to the mass of the gaseous products lost.
Step 3: Detailed Explanation:
First, calculate the molar masses of the relevant compounds:
- Molar mass of NaHCO₃ = 23 + 1 + 12 + 3(16) = 84 g/mol.
- Molar mass of H₂O = 2(1) + 16 = 18 g/mol.
- Molar mass of CO₂ = 12 + 2(16) = 44 g/mol.
From the balanced equation, the decomposition of 2 moles of NaHCO₃ results in the loss of 1 mole of H₂O and 1 mole of CO₂.
- Mass of 2 moles of NaHCO₃ = \(2 \times 84 = 168\) g.
- Total mass of gaseous products lost = Mass of H₂O + Mass of CO₂ = \(18 + 44 = 62\) g.
This means that 168 g of NaHCO₃ produces a mass loss of 62 g.
We are given that the actual loss in mass is 0.62 g. We can set up a proportion to find the mass of NaHCO₃ in the original mixture.
Let \(x\) be the mass of NaHCO₃ in the mixture.
\[ \frac{\text{Mass of NaHCO}_3}{\text{Mass loss}} = \frac{168 \text{ g}}{62 \text{ g}} \] \[ \frac{x}{0.62 \text{ g}} = \frac{168 \text{ g}}{62 \text{ g}} \] \[ x = \frac{168}{62} \times 0.62 = 1.68 \text{ g} \] So, the mass of NaHCO₃ in the mixture is 1.68 g.
The total mass of the mixture is 4.0 g.
Mass of Na₂CO₃ = Total mass - Mass of NaHCO₃ = \(4.0 \text{ g} - 1.68 \text{ g} = 2.32 \text{ g}\).
Finally, calculate the percentage of sodium carbonate (Na₂CO₃) in the mixture:
\[ % \text{ Na}_2\text{CO}_3 = \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Total mass of mixture}} \times 100 \] \[ % \text{ Na}_2\text{CO}_3 = \frac{2.32 \text{ g}}{4.0 \text{ g}} \times 100 = 0.58 \times 100 = 58 % \] Step 4: Final Answer:
The percentage of sodium carbonate in the mixture is 58%. Therefore, option (B) is correct.