NECO 2009 Chemistry · Question 8

What volume of carbon (IV) oxide in dm 3 is produced at s.t.p. when 2.50g of CaCO3 reacts with excess acid…

Question

What volume of carbon (IV) oxide in dm 3 is produced at s.t.p. when 2.50g of CaCO3 reacts with excess acid according to the following equation? CaCO3(s) + 2HCI(aq) → CaCl2(aq) + H2O(1) + CO2(g) [CaCO2 100; molar volume of a gas at s.t.p. = 22.4dm 3]

A.0.28
B.0.56
C.2.24
D.5.6
E.11.2

Answer: B. 0.56

Explanation written with AI and checked against the answer key

Analyze the question: We need the volume of CO₂ released at s.t.p. when 2.50 g of CaCO₃ reacts with excess HCl. Concept: From the balanced equation, 1 mole of CaCO₃ gives 1 mole of CO₂. At s.t.p., 1 mole of any gas occupies 22.4 dm³. Solution: Step 1: Find the moles of CaCO₃. Molar mass of CaCO₃ = 100 g/mol (given) Moles = mass ÷ molar mass = 2.50 ÷ 100 = 0.025 mol Step 2: Use the mole ratio. CaCO₃ : CO₂ = 1 : 1, so moles of CO₂ = 0.025 mol Step 3: Convert to volume. Volume = 0.025 × 22.4 = 0.56 dm³ Answer: The volume of CO₂ produced is 0.56 dm³, which is option B. Tip: For this type of question, follow the chain mass → moles → mole ratio → volume. Note that 2.50 g is 1/40 of 100 g, so the answer is 22.4 ÷ 40 = 0.56. Spotting a simple fraction like this saves time in the exam.

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