Step 1: Understand the equilibrium system.
The given reaction is a heterogeneous equilibrium:
\[
CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)
\]
Here solids have constant activity, so only gaseous $CO_2$ affects equilibrium position. The system shifts according to Le Chatelier’s principle depending on pressure, temperature, or removal/addition of gaseous species.
Step 2: Effect of adding solid reactant.
Adding more $CaCO_3$ does not affect equilibrium because pure solids do not appear in equilibrium constant expression. Their activity is constant and independent of amount. Hence, loading more solid does not shift equilibrium.
Step 3: Effect of pressure increase.
Increasing pressure shifts equilibrium towards the side with fewer moles of gas. Left side has zero gas, right side has 1 mole of gas ($CO_2$). So increasing pressure shifts equilibrium to the left, preventing decomposition rather than completing it.
Step 4: Effect of temperature.
The decomposition of $CaCO_3$ is endothermic. Increasing temperature favors product formation, but does not guarantee complete conversion unless CO$_2$ is removed continuously. So temperature alone is insufficient.
Step 5: Effect of removing CO$_2$.
If $CO_2$ is continuously removed, its partial pressure remains low. This drives equilibrium forward continuously to produce more $CO_2$, leading to complete decomposition of $CaCO_3$. This is the most effective condition.
Step 6: Final conclusion.
Thus, complete conversion is achieved by:
\[
\boxed{\text{Removing CO}_2 \text{ continuously}}
\]
