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Equilibrium constant $K_{p}$ for the reaction
${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g) is 0.82 atm at 727 ° C$
If 1 mole of ${CaCO}_{3}$ is placed in a closed container of 20L and heated to this temperature, what amount of ${CaCO}_{3}$ in grams would dissociate at equilibrium ?

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Detailed Solution

${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

$K_{p} = P_{{CO}_{2}} = 0.82 atm$

$n_{{CO}_{2}} = \frac{PV}{RT} = \frac{0.82 \times 20}{0.082 \times 1000} = 0.2 mole$

$Mole of {CaCO}_{3} dissociated = n_{{CO}_{2}} = 0.2$

$Amount dissociated = 0.2 \times 100 = 20 g$

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