To solve this problem, we need to determine the mass of $$CO$_2$ produced by heating 20 g of 20% pure limestone.$\\$Given:$\\$• The limestone is 20% pure, which means it contains 20% $$CaCO$_3. \\$• Mass of limestone $= 20$ g$\\$• Purity of limestone $= 20\% \\$Step 1: Calculate the mass of pure $$CaCO$_3$ in the limestone.$\\$Mass of pure $$CaCO$_3 = \frac{20}{100} \times 20 \\ = 4$ g$\\$Step 2: Write the decomposition reaction of $$CaCO$_3. \\$CaCO$_3 \xrightarrow{1200 K}$CaO$+$CO$_2 \\$Step 3: Calculate the molar mass of $$CaCO$_3$ and $$CO$_2. \\$Molar mass of $$CaCO$_3 = 40 ($Ca$) + 12 ($C$) + 3 \times 16 ($O$) \\ = 100$ g/mol$\\$Molar mass of $$CO$_2 = 12 ($C$) + 2 \times 16 ($O$) \\ = 44$ g/mol$\\$Step 4: Use stoichiometry to find the mass of $$CO$_2$ produced.$\\$From the balanced equation, 1 mole of $$CaCO$_3$ produces 1 mole of $$CO$_2. \\$Therefore, 100 g of $$CaCO$_3$ produces 44 g of $$CO$_2. \\$Step 5: Calculate the mass of $$CO$_2$ produced from 4 g of $$CaCO$_3. \\$Using the ratio:$\newline \frac{44 g \ of \ {CO}_2}{100 g \ of \ {CaCO}_3} = \frac{x \ g of \ {CO}_2}{4 \ g of \ {CaCO}_3} \\$Solve for $x: \\ x = \frac{44}{100} \times 4 \\ = 1.76$ g$\\$Therefore, the mass of $$CO$_2$ produced is $1.76$ g.$\\$This corresponds to Option 2.