The mass of ${\text{CO}}_2$ produced by heating $20\text{g}$ of $20\%$ pure limestone is:(Atomic mass of $\text{Ca}=40$)
$${\text{CaCO}}_3\stackrel{1200\text{K}}{\to }\text{CaO}+{\text{CO}}_2$$

Step 1: Find the mass of pure ${\text{CaCO}}_3$:Pure mass $=20\%\text{ of }20\text{g}=0.20\times 20=4\text{g}$.Step 2: Calculate moles of ${\text{CaCO}}_3$:Molar mass of ${\text{CaCO}}_3=40+12+(3\times 16)=100\text{g/mol}$.Moles $=\frac{4}{100}=0.04\text{mol}$.Step 3: From stoichiometry, $1$ mole ${\text{CaCO}}_3$ yields $1$ mole ${\text{CO}}_2$.Moles of ${\text{CO}}_2$ produced $=0.04\text{mol}$.Step 4: Convert to mass:Molar mass of ${\text{CO}}_2=12+(2\times 16)=44\text{g/mol}$.Mass $=0.04\times 44=1.76\text{g}$.