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JEE MainChemistryToughPhysical Chemistry

What is the minimum mass of CaCO_3 (s), below which it
decomposes completely, required to establish equilibrium in a 6.50 litre
container for the reaction:
CaCO3(s)⇌CaO(s)+CO2(g){CaCO}_{3}(s) \rightleftharpoons CaO(s) + {CO}_{2}(g); K_c = 0.05

Show answer and solution
K_C = [CO_2] = 0.05 mole/litre
so moles of CO_2 = 6.50 × 0.05 moles = 0.3250 moles
CaCO3(s)⇌CaO(s)+CO2(g){CaCO}_{3}(s) \rightleftharpoons CaO(s) + {CO}_{2}(g)
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 g of
CaCO_3 = 32.5 gm of CaCO_3

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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