CH₃CH₂COO– Na+ $\stackrel{NaOH,+?}{\underset{Heat}{⟶<!--\; ⟶\; -->}}$CH₃CH₃ + Na₂CO₃.

Consider the above reaction and identify the missing reagent/chemical.

The correct option is CaO.

Step-by-step Explanation:

1. Analyze the Reactants and Products:
The given chemical equation is:

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{COO}}^{-}{\mathrm{Na}}^{+}\underset{\mathrm{Heat}}{\overset{\mathrm{NaOH}+?}{\to }}{\mathrm{CH}}_3{\mathrm{CH}}_3+{\mathrm{Na}}_2{\mathrm{CO}}_3$

Here, the reactant is sodium propanoate (a sodium salt of a carboxylic acid) and the organic product is ethane (an alkane containing one less carbon atom than the parent carboxylic acid salt), along with sodium carbonate (${\mathrm{Na}}_2$).

2. Identify the Reaction Type:
The conversion of a sodium salt of a carboxylic acid to an alkane with the elimination of carbon dioxide (in the form of carbonate) is known as decarboxylation (specifically, soda-lime decarboxylation).

3. Determine the Missing Reagent:
Decarboxylation of sodium salts of carboxylic acids is carried out by heating the salt with soda-lime.
Soda-lime is a mixture of sodium hydroxide ($\mathrm{NaOH}$) and calcium oxide (CaO) in a ratio of approximately 3:1.
Therefore, the missing reagent/chemical marked by '$?$' in the presence of $\mathrm{NaOH}$ and heat is CaO (calcium oxide).