An antacid tablet is known to contain calcium carbonate (CaCO₃) - HSC - SSCE Chemistry - Question 24 - 2005 - Paper 1

To calculate the moles of hydrochloric acid added, use the formula:

$\text{Moles} = \text{Concentration} \times \text{Volume}$

Given:

• Concentration of HCl = 0.600 mol L⁻¹
• Volume of HCl = 25.0 mL = 0.0250 L

Thus,

$\text{Moles of HCl} = 0.600 \text{ mol L⁻¹} \times 0.0250 \text{ L} = 0.0150 \text{ mol}$

First, we need to find the moles of excess HCl that were neutralized by sodium hydroxide. Since we have 14.2 mL of NaOH at a concentration of 0.100 mol L⁻¹:

1. Calculate moles of NaOH:

$\text{Moles of NaOH} = 0.100 \text{ mol L⁻¹} \times 0.0142 \text{ L} = 0.00142 \text{ mol}$

2. Since NaOH reacts with HCl in a 1:1 molar ratio, the moles of HCl neutralized by NaOH = 0.00142 mol.

3. Calculate moles of HCl that reacted with CaCO₃:

$\text{Moles of HCl reacted} = 0.0150 - 0.00142 = 0.01358 \text{ mol}$

4. From the balanced equation, 1 mol of CaCO₃ reacts with 2 mol of HCl. So, moles of CaCO₃ = 0.01358 mol / 2 = 0.00679 mol.

5. Now, calculate the mass of CaCO₃:

$\text{Mass} = \text{Moles} \times \text{Molar mass (CaCO₃)}$

Molar mass of CaCO₃ = 100.09 g/mol,

$\text{Mass of CaCO₃} = 0.00679 \text{ mol} \times 100.09 \text{ g/mol} \approx 0.679 \text{ g}$