Question

Percentage loss in mass on heating mixture of $Na_2CO_3$ & $CaCO_3$ containing equal mass of the two components will be

• A. 44%
• B. 22%
• C. 35%
• D. 50%

Answer 1

Answer: (B)

22%

Answer 2

To determine the percentage loss in mass on heating a mixture of $Na_2CO_3$ and $CaCO_3$ containing equal mass of the two components, we need to consider the thermal stability of each compound.

1. Thermal Decomposition of $Na_2CO_3$:
   Sodium carbonate ($Na_2CO_3$) is thermally stable and does not decompose on heating under normal laboratory conditions (i.e., it does not decompose to release $CO_2$ at temperatures where $CaCO_3$ decomposes). Therefore, there will be no loss of mass from $Na_2CO_3$.

2. Thermal Decomposition of $CaCO_3$:
   Calcium carbonate ($CaCO_3$) decomposes on heating to form calcium oxide ($CaO$) and carbon dioxide ($CO_2$).
   The reaction is:
   $CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$

   To calculate the mass loss, we need the molar masses:

   • Molar mass of $CaCO_3 = 40.08 (Ca) + 12.01 (C) + 3 \times 16.00 (O) = 100.09 \text{ g/mol}$ (approximately $100 \text{ g/mol}$)
   • Molar mass of $CO_2 = 12.01 (C) + 2 \times 16.00 (O) = 44.01 \text{ g/mol}$ (approximately $44 \text{ g/mol}$)

   From the balanced equation, 1 mole of $CaCO_3$ (100.09 g) produces 1 mole of $CO_2$ (44.01 g). The mass loss is due to the evolution of $CO_2$.

3. Calculation for the Mixture:
   Let the mass of $Na_2CO_3$ in the mixture be 'm' grams.
   Since the mixture contains equal mass of the two components, the mass of $CaCO_3$ will also be 'm' grams.
   Total initial mass of the mixture = $m (\text{Na}_2\text{CO}_3) + m (\text{CaCO}_3) = 2m$ grams.

   Upon heating:

   • Mass loss from $Na_2CO_3 = 0$ grams.
   • Mass loss from $CaCO_3$:
     For every $100.09 \text{ g}$ of $CaCO_3$, $44.01 \text{ g}$ of $CO_2$ is lost.
     So, for 'm' grams of $CaCO_3$, the mass lost will be:
     Mass loss = $\frac{44.01 \text{ g } CO_2}{100.09 \text{ g } CaCO_3} \times m \text{ g } CaCO_3 \approx \frac{44}{100} \times m = 0.44m$ grams.

   Total mass lost from the mixture = $0 + 0.44m = 0.44m$ grams.

4. Percentage Loss in Mass:
   Percentage loss in mass = $\frac{\text{Total mass lost}}{\text{Total initial mass}} \times 100\%$
   Percentage loss = $\frac{0.44m}{2m} \times 100\%$
   Percentage loss = $\frac{0.44}{2} \times 100\%$
   Percentage loss = $0.22 \times 100\%$
   Percentage loss = $22\%$