The reaction described is the decarboxylation of sodium ethanoate using soda lime (a mixture of sodium hydroxide and calcium oxide) [Source 180]. $CH_3COO^-Na^+ + NaOH \xrightarrow{CaO, \ \Delta} CH_4 + Na_2CO_3$ The organic compound formed in this reaction is methane ($CH_4$). The molar mass of methane ($CH_4$) is calculated as: $12 + 4(1) = 16 \text{ g/mol}$. Therefore, the weight of two moles of methane is: $2 \text{ moles} \times 16 \text{ g/mol} = 32 \text{ g}$.