The right option for the mass of ${\mathrm{CO}}_2$ produced by heating $20 \mathrm{g}$ of $20\%$ pure limestone is (Atomic mass of $\mathrm{Ca}=40$)

$\left[{\mathrm{CaCO}}_3\stackrel{1200 \mathrm{K}}{\to }\mathrm{CaO}+{\mathrm{CO}}_2\right]$

${\mathrm{CaCO}}_3\stackrel{1200 \mathrm{K}}{\to }\mathrm{CaO}+{\mathrm{CO}}_2$

According to the given equation, 100 grams of calcium carbonate gives 44 grams of carbon dioxide. 

Given that 20 grams of 20% pure lime stone, it means the given lime stone contain 4 grams of ${\mathrm{CaCO}}_3$.

Hence, the maximum amount of ${\mathrm{CO}}_2$ produced 

$=\frac{4\times 44}{100}=1.76 \mathrm{g}$