Chemical EquilibriumHard
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) ⇋ CaO(s) + CO2(g).
Options
A.32.5 g
B.24.6 g
C.40.9 g
D.8.0 gm
Solution
KC = [CO2] = 0.05 mole/litre
so moles of CO2 = 6.50 × 0.05 moles = 0.3250 moles
CaCO3 ⇋ 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
so moles of CO2 = 6.50 × 0.05 moles = 0.3250 moles
CaCO3 ⇋ 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
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