What mass of 95\% pure CaCO_3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO_3(s) + 2HCl(aq) → C - Chemistry MCQ NEET 2022 2026 | ExamTest.live

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What mass of 95% pure CaCO_{3} will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO_{3} (s) + 2 HCl (a q) \to CaCl_{2} (a q) + CO_{2} (g) + 2 H_{2} O (l)

Solution & Step-by-step Explanation

Moles of HCl = Molarity \times Volume (L) = 0.5 \times 0.050 = 0.025 mol .From the balanced equation, 2 moles of HCl react with 1 mole of CaCO_{3} .Moles of CaCO_{3} required = \frac{0.025}{2} = 0.0125 mol .Molar mass of CaCO_{3} = 100 g/mol .Pure mass of CaCO_{3} required = 0.0125 \times 100 = 1.25 g .Since the sample is only 95% pure, let total mass be x : 0.95 \times x = 1.25 ⟹ x = \frac{1.25}{0.95} \approx 1.3157 g \approx 1.32 g

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What mass of 95% pure CaCO_{3} will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO_{3} (s) + 2 HCl (a q) \to CaCl_{2} (a q) + CO_{2} (g) + 2 H_{2} O (l)

1.25 g

1.32 g

3.65 g

9.50 g

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