What mass of 95% pure CaCO₃ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
CaCO_3(s)+2HCl_(aq)→CaCl_2(aq)+CO_2(g)+2H2O₍₁₎
[Calculate up to second place of decimal point]

To determine the mass of 95% pure CaCO₃ required to neutralize 50 mL of 0.5 M HCl solution, we will follow a step-by-step calculation based on the given balanced chemical equation:

Balanced chemical equation:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + 2H₂O(l)

The first step is to calculate the number of moles of HCl in the given solution. We know that:

• Volume of HCl solution = 50 mL = 0.050 L
• Concentration of HCl = 0.5 M

Number of moles of HCl = Concentration × Volume = 0.5 M × 0.050 L = 0.025 moles.

According to the balanced chemical equation, 2 moles of HCl neutralize 1 mole of CaCO₃. So, the moles of CaCO₃ required will be:

Moles of CaCO₃ = (0.025 moles of HCl) × (1 mole of CaCO₃ / 2 moles of HCl) = 0.0125 moles.

Now, we calculate the mass of CaCO₃ required. The molar mass of CaCO₃ is approximately 100 g/mol. Therefore, the mass of 100% pure CaCO₃ required will be:

Mass of CaCO₃ = Moles × Molar Mass = 0.0125 moles × 100 g/mol = 1.25 g.

Since the CaCO₃ available is only 95% pure, we need to adjust the mass calculation for the purity. The mass of 95% pure CaCO₃ required is:

Mass of 95% pure CaCO₃ = 1.25 g / 0.95 = 1.32 g.

Therefore, the correct answer is: 1.32 g