Question

Determine the percentage composition (by mass) of a mixture of anhydrous sodium carbonate and sodium bicarbonate from the following data:

wt. of the mixture taken = 2g Loss in weight on heating = 0.124 gm.

Answer 1

Sodium bicarbonate (NaHCO3): 16.8% Sodium carbonate (Na2CO3): 83.2%

Answer 2

The problem involves the thermal decomposition of a mixture containing anhydrous sodium carbonate ($\text{Na}_2\text{CO}_3$) and sodium bicarbonate ($\text{NaHCO}_3$).

1. Identify the decomposing component: Anhydrous sodium carbonate ($\text{Na}_2\text{CO}_3$) is thermally stable and does not decompose on heating. Sodium bicarbonate ($\text{NaHCO}_3$) decomposes on heating, leading to a loss in mass.

2. Write the balanced chemical equation for decomposition: The decomposition of sodium bicarbonate is given by: $2\text{NaHCO}_3(\text{s}) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{g}) + \text{CO}_2(\text{g})$

3. Calculate molar masses:

   • Molar mass of $\text{NaHCO}_3 = 23 (\text{Na}) + 1 (\text{H}) + 12 (\text{C}) + 3 \times 16 (\text{O}) = 84 \text{ g/mol}$
   • Molar mass of $\text{H}_2\text{O} = 2 \times 1 (\text{H}) + 16 (\text{O}) = 18 \text{ g/mol}$
   • Molar mass of $\text{CO}_2 = 12 (\text{C}) + 2 \times 16 (\text{O}) = 44 \text{ g/mol}$

4. Relate mass of $\text{NaHCO}_3$ to mass loss: From the balanced equation, 2 moles of $\text{NaHCO}_3$ (i.e., $2 \times 84 = 168$ g) decompose to produce 1 mole of $\text{H}_2\text{O}$ (18 g) and 1 mole of $\text{CO}_2$ (44 g). The total mass loss from 168 g of $\text{NaHCO}_3$ is $18 \text{ g} (\text{H}_2\text{O}) + 44 \text{ g} (\text{CO}_2) = 62 \text{ g}$.

5. Calculate the mass of $\text{NaHCO}_3$ in the mixture: Let the mass of $\text{NaHCO}_3$ in the mixture be $x$ g. We know that 168 g of $\text{NaHCO}_3$ causes a mass loss of 62 g. Given loss in weight on heating = 0.124 g. Using proportionality: $\frac{x \text{ g NaHCO}_3}{0.124 \text{ g (mass loss)}} = \frac{168 \text{ g NaHCO}_3}{62 \text{ g (mass loss)}}$ $x = 0.124 \times \frac{168}{62}$ $x = 0.002 \times 168$ $x = 0.336 \text{ g}$ So, the mass of sodium bicarbonate ($\text{NaHCO}_3$) in the mixture is 0.336 g.

6. Calculate the mass of $\text{Na}_2\text{CO}_3$ in the mixture: Total mass of the mixture = 2 g. Mass of $\text{Na}_2\text{CO}_3 = \text{Total mass of mixture} - \text{Mass of NaHCO}_3$ Mass of $\text{Na}_2\text{CO}_3 = 2 \text{ g} - 0.336 \text{ g}$ Mass of $\text{Na}_2\text{CO}_3 = 1.664 \text{ g}$

7. Determine the percentage composition by mass: Percentage of $\text{NaHCO}_3 = \frac{\text{Mass of NaHCO}_3}{\text{Total mass of mixture}} \times 100\%$ Percentage of $\text{NaHCO}_3 = \frac{0.336 \text{ g}}{2 \text{ g}} \times 100\% = 16.8\%$

   Percentage of $\text{Na}_2\text{CO}_3 = \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Total mass of mixture}} \times 100\%$ Percentage of $\text{Na}_2\text{CO}_3 = \frac{1.664 \text{ g}}{2 \text{ g}} \times 100\% = 83.2\%$

The percentage composition of the mixture is 16.8% sodium bicarbonate and 83.2% sodium carbonate.

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Explanation of the solution:

1. Sodium bicarbonate ($\text{NaHCO}_3$) decomposes on heating, while sodium carbonate ($\text{Na}_2\text{CO}_3$) does not.
2. The decomposition reaction is $2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$.
3. Molar mass of $\text{NaHCO}_3$ is 84 g/mol. Molar mass of $\text{H}_2\text{O}$ is 18 g/mol, and $\text{CO}_2$ is 44 g/mol.
4. From stoichiometry, 168 g ($2 \times 84$) of $\text{NaHCO}_3$ produces a mass loss of 62 g ($18 + 44$).
5. Given loss in weight is 0.124 g.
6. Mass of $\text{NaHCO}_3 = 0.124 \text{ g} \times \frac{168 \text{ g}}{62 \text{ g}} = 0.336 \text{ g}$.
7. Mass of $\text{Na}_2\text{CO}_3 = \text{Total mass} - \text{Mass of NaHCO}_3 = 2 \text{ g} - 0.336 \text{ g} = 1.664 \text{ g}$.
8. Percentage of $\text{NaHCO}_3 = \frac{0.336}{2} \times 100\% = 16.8\%$.
9. Percentage of $\text{Na}_2\text{CO}_3 = \frac{1.664}{2} \times 100\% = 83.2\%$.