The calcium carbonate (CaCO3) in antacid tablets reacts with dilute hydrochloric acid (HCl) according to the following balanced equation: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) ΔH < 0 5.1 Is the above reaction EXOTHERMIC or ENDOTHERMIC? Give a reason for the answer - NSC Physical Sciences - Question 5 - 2019 - Paper 2

The average rate can be calculated using the formula:

$\text{Rate} = \frac{\Delta m}{\Delta t}$

The mass lost is 2 g - 0.25 g = 1.75 g and the time taken is 30 s.

Thus,

$\text{Rate} = \frac{1.75 \text{ g}}{30 \text{ s}} = 0.0583 \text{ g s}^{-1}$.

First, calculate the mass of calcium carbonate present in the tablet:

Mass of antacid tablet = 2 g Percentage of CaCO3 = 40%

Mass of CaCO3 = 2 g × 0.4 = 0.8 g.

Now, convert grams of CaCO3 to moles:

Molar mass of CaCO3 = 100 g/mol,

Moles of CaCO3 = 0.8 g / 100 g/mol = 0.008 mol.

According to the balanced equation:

1 mol of CaCO3 produces 1 mol of CO2.

Thus, 0.008 mol of CaCO3 produces 0.008 mol of CO2.

At STP, 1 mole of gas occupies 22.4 dm³;

Volume of CO2 = 0.008 mol × 22.4 dm³/mol = 0.1792 dm³.