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Equilibrium constant K p for the reaction CaCO 3 ⇌ CaO + CO 2 is 0.82 atm at 7 2 7 ∘ C. If 1 mole of CaCO 3 is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO 3 would dissociate at equilibrium?

Options

  1. A0.2 g
  2. B80 g
  3. C20 g
  4. D50 g

Correct answer

C. 20 g

Step-by-step solution

CaCO 3 ⇌ CaO + CO 2 K p = p CO 2 = 0.82 atm 0.82 atm × 20 L = n CO 2 × 0.082 × 1000 2 0 0 1 0 0 0 = 1 5 = n CO 2 No. of moles CO 2 = no. of moles of CaCO 3 decomposed = 1 5 mole Amount of CaCO 3 decomposed = 1 5 × 100 = 20 g

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