Question

What is the minimum mass of CaCO3 (s), below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction : CaCO3(s) $\rightleftharpoons$

CaO(s) + CO2(g); Kc = 0.05 mole/litre

• A. 32.5 g
• B. 24.6 g
• C. 40.9 g
• D. 8.0 gm

Answer 1

Answer: (A)

32.5 g

Answer 2

KC = [CO2] = 0.05 mole/litre

so moles of CO2 = 6.50 × 0.05 moles = 0.3250 moles

CaCO3 $\rightleftharpoons$ CaO + CO2

1 mole of CO2 = 1 mole of CaCO3

0.3250 moles of CO2 = 0.3250 moles of CaCO3

= 0.3250 × 100 gm of CaCO3 = 32.5 gm of CaCO3