What mass of CaCO3 gives 48.8 litres (measured at STP) of CO2, on strong heating?_.
Correct answer: C. 200g
- A. 100g
- B. 150g
- C. 200g
- D. 300g
Explanation
Write the balanced chemical equation: CaCO3(s) -> CaO(s) + CO2(g)Calculate the moles of CO2:At STP, 1 mole of any gas occupies 22.4 liters.So, moles of CO2 = 48.8 L / 22.4 L/mol = 2.18 molesUse the mole ratio from the balanced equation:1 mole of CaCO3 produces 1 mole of CO2. So, moles of CaCO3 = 2.18 molesCalculate the mass of CaCO3:Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 100.09 g/mol Mass of CaCO3 = moles * molar mass = 2.18 moles * 100.09 g/mol = 218.2 gTherefore, the mass of CaCO3 that gives 48.8 liters of CO2 at STP is approximately 218.2 grams.
This question appeared on the KMU MDCAT 2023 (retake) paper, which you can sit online with every answer explained.
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