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CGP EDU Academic Team
Published on: August 14, 2026
What is the minimum mass of CaCO 3 (s), below which it decomposes completely, required to establish equilibrium in a 6.50 liter container for the reaction: CaCO 3 (s) 
CaO(s) + CO 2 (g); K c = 0.05 mole/liter
Text Solution
Verified by ExpertsThe correct answer is:
A
K C = [CO 2 ] = 0.05 mole/liter
so moles of CO 2 = 6.50 × 0.05 moles = 0.3250 moles
CaCO 3 
CaO + CO 2
1 mole of CO 2 = 1 mole of CaCO 3
0.3250 moles of CO 2 = 0.3250 moles of CaCO 3
= 0.3250 × 100 gm of CaCO 3 = 32.5 gm of CaCO 3
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