Chemistry · Some Basic Concepts Of Chemistry · NEET
Methane gas (CH4) is formed. The full reaction is CH3COONa + NaOH --(CaO, heat)--> CH4 + Na2CO3. The sodium acetate loses its carboxylate part as sodium carbonate (Na2CO3), and the leftover CH3 group picks up a hydrogen to become CH4. So a 2-carbon salt gives a 1-carbon gas.
Soda lime is a MIXTURE of two things: sodium hydroxide (NaOH) and calcium oxide (CaO). NaOH is the part that actually reacts with the sodium acetate. CaO is added as a helper. In NEET questions the reagent may be written as 'NaOH in presence of CaO' or simply 'soda lime' - both mean the same thing.
Pure molten NaOH is sticky, corrosive, and it absorbs water and CO2 from the air, so it attacks glass and gets spoiled. Mixing solid CaO with NaOH makes soda lime a dry, easy-to-handle solid that does not melt and stick. CaO also keeps the mixture dry by soaking up moisture. The chemistry is done by NaOH; CaO is just a physical helper, so you do NOT write CaO in the balanced equation - it sits above the arrow as a condition.
Decarboxylation means removing the carboxyl group (-COOH or its salt -COONa) from a molecule and losing it as carbon dioxide (or as carbonate). The molecule ends up with one fewer carbon atom. Here CH3COONa loses -COONa and becomes CH4. This is the standard lab method to make methane (or to take any carboxylic acid salt down by one carbon).
It gives methane (CH4) plus sodium carbonate (Na2CO3), but only when heated in the presence of CaO (soda lime). Without heat and CaO, mixing sodium acetate and NaOH does nothing useful. Remember the conditions: heat + soda lime are required.
Yes. Ethanoate is the IUPAC name and acetate is the common name for the same ion, CH3COO-. So 'sodium ethanoate' = 'sodium acetate' = CH3COONa. NEET 2023 wrote it as 'sodium ethanoate' to test if you know this. Both heated with NaOH and CaO give methane.
Methane CH4 has molar mass = 12 + (4 x 1) = 16 g/mol. NEET links this reaction to mole-mass math: they ask for the mass of methane formed. For example, 2 mol of CH4 = 2 x 16 = 32 g. Always convert to grams using 16 g/mol.
The weight (in g) of two moles of the organic compound obtained by heating sodium ethanoate with sodium hydroxide in the presence of calcium oxide is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
CH3COONa + NaOH --(CaO, heat)--> CH4 + Na2CO3. Methane is the gas; sodium carbonate stays behind. CaO is a condition, not a reactant with a coefficient.
Because NaOH is a strong base. The CO2 that would form is immediately trapped by the excess NaOH to give sodium carbonate (Na2CO3). So in this reaction you collect only methane as the gas, and the carbonate remains in the solid residue.
Yes - the same soda-lime decarboxylation works on any sodium salt of a carboxylic acid and removes one carbon. Sodium propanoate (CH3CH2COONa) gives ethane (C2H6). The general rule: R-COONa gives R-H. Acetate gives methane because R = CH3 becomes CH4.
Exactly 1 mole of methane, because the mole ratio is 1:1. So 1 mole CH3COONa gives 1 mole CH4 = 16 g. This 1:1 ratio is the key to any NEET numerical on this reaction.
