1 Explain why the reaction is exothermic. 15.2 Calculate the average rate of reaction. 15.3 Calculate the moles of calcium carbonate in the tablet.

$$\text{Change in mass} = \text{Initial mass} - \text{Final mass}$$ $$\text{Change in mass} = 2 \text{ g} - 0.25 \text{ g} = 1.75 \text{ g}$$

$$\text{Average rate} = \frac{\text{Change in mass}}{\text{Time taken}}$$ $$\text{Average rate} = \frac{1.75 \text{ g}}{30 \text{ s}}$$ $$\text{Average rate} = 0.05833... \text{ g} \cdot \text{s}^{-1}$$ $$\text{Average rate} = \boxed{\text{0.058 g} \cdot \text{s}^{-1}}$$

$$\text{Mass of CaCO}_3 = 40\% \text{ of } 2 \text{ g}$$ $$\text{Mass of CaCO}_3 = 0.40 \times 2 \text{ g} = 0.80 \text{ g}$$

$$\text{M}(\text{CaCO}_3) = \text{M}(\text{Ca}) + \text{M}(\text{C}) + 3 \times \text{M}(\text{O})$$ $$\text{M}(\text{CaCO}_3) = 40.08 \text{ g/mol} + 12.01 \text{ g/mol} + 3 \times 16.00 \text{ g/mol}$$ $$\text{M}(\text{CaCO}_3) = 100.09 \text{ g/mol}$$

$$\text{Moles of CaCO}_3 = \frac{\text{Mass of CaCO}_3}{\text{Molar mass of CaCO}_3}$$ $$\text{Moles of CaCO}_3 = \frac{0.80 \text{ g}}{100.09 \text{ g/mol}} = 0.0079928... \text{ mol}$$

The balanced equation is: $\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \to \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$
From the equation, 1 mole of $\text{CaCO}_3$ produces 1 mole of $\text{CO}_2$.

$$\text{Moles of CO}_2 = \text{Moles of CaCO}_3 = 0.0079928... \text{ mol}$$

At STP, 1 mole of any gas occupies $22.4 \text{ dm}^3$.

$$\text{Volume of CO}_2 = \text{Moles of CO}_2 \times 22.4 \text{ dm}^3/\text{mol}$$ $$\text{Volume of CO}_2 = 0.0079928... \text{ mol} \times 22.4 \text{ dm}^3/\text{mol}$$ $$\text{Volume of CO}_2 = 0.179038... \text{ dm}^3$$ $$\text{Volume of CO}_2 = \boxed{\text{0.179 dm}^3}$$