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: Decomposition Reaction: Only \( \text{NaHCO}_3 \) decomposes. \[ 2\text{NaHCO}_3 \to \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \] Mass loss is due to \( \text{H}_2\text{O} \) and \( \text{CO}_2 \).
Step 2: Stoichiometry: 2 moles of \( \text{NaHCO}_3 \) ($2 \times 84 = 168$ g) produce 1 mole \( \text{H}_2\text{O} \) (18g) + 1 mole \( \text{CO}_2 \) (44g). Total mass loss for 168 g reactant = \( 18 + 44 = 62 \) g.
Step 3: Calculate Mass of \( \text{NaHCO}_3 \): Given loss = 0.62 g. Mass of \( \text{NaHCO}_3 \) = \( \frac{168}{62} \times 0.62 = 1.68 \) g.
Step 4: Percentage of \( \text{Na}_2\text{CO}_3 \): Total mass = 4.0 g. Mass of \( \text{Na}_2\text{CO}_3 \) = \( 4.0 - 1.68 = 2.32 \) g. Percentage = \( \frac{2.32}{4.0} \times 100 = 58% \).