$0.5 \ mol$ of $CaCO_3(s)$ is heated in a $500 \ mL$ closed vessel at $400 \ K$. The reaction is $CaCO_{3(s)} \rightleftharpoons CaO(s) + CO_{2(g)}$. If the equilibrium constant $K_c = 0.9 \ mol \ L^{-1}$,calculate the amount of $CO_2$ in $mol$ at equilibrium and the percentage of the reaction completed.

(A) The equilibrium constant expression for the reaction is $K_c = [CO_2]$.
Given $K_c = 0.9 \ mol \ L^{-1}$,therefore at equilibrium,$[CO_2] = 0.9 \ mol \ L^{-1}$.
The volume of the vessel is $500 \ mL = 0.5 \ L$.
The amount of $CO_2$ at equilibrium is $n(CO_2) = [CO_2] \times V = 0.9 \ mol \ L^{-1} \times 0.5 \ L = 0.45 \ mol$.
Since $1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$,the amount of $CaCO_3$ decomposed is $0.45 \ mol$.
The initial amount of $CaCO_3$ is $0.5 \ mol$.
Percentage of reaction completed = $\frac{0.45 \ mol}{0.5 \ mol} \times 100 = 90 \%$.