To solve this problem, we need to understand the reaction involved when sodium ethanoate is heated with sodium hydroxide in the presence of calcium oxide. This reaction is known as the decarboxylation reaction, which produces methane.$\\$The chemical equation for the reaction is:$\\ {CH}_3COONa + NaOH \xrightarrow{CaO, heat}$CH$_4 +$Na$_2$CO$_3 \\$In this reaction, sodium ethanoate $({CH}_3COONa)$ reacts with sodium hydroxide (NaOH) to produce methane$({CH}_4)$ and sodium carbonate $({Na}_2{CO}_3). \\$Let's analyze the stoichiometry of the reaction:$\\$• One mole of sodium ethanoate produces one mole of methane.$\\$Therefore, two moles of sodium ethanoate will produce two moles of methane.$\\$Next, we calculate the weight of two moles of methane:$\\$• The molecular weight of methane $(CH_4)$is calculated as follows:$\\$• Carbon (C) has an atomic weight of 12 g/mol.$\\$• Hydrogen (H) has an atomic weight of 1 g/mol, and there are 4 hydrogen atoms in methane.$\\$Thus, the molecular weight of methane is:$\\ 12 + (4 \times 1) = 16$ g/mol$\\$Therefore, the weight of two moles of methane is:$\\ 2 \times 16 = 32$ g$\\$Hence, the weight of two moles of the organic compound (methane) obtained is 32 g.$\\$This corresponds to Option 2.