The right option for the mass of${\text{CO}}_{\text{2}}$Produced by heating of 50g of 50% pure lime stone is (Atomic mass of Ca = 40)
$[{\text{CaCO}}_{\text{3}}\stackrel{\text{1200\hspace{0.17em}K}}{\to }\text{CaO}+{\text{CO}}_2]$

Mass of pure CaCO₃ = 50 × 50 /100 = 25g
${\text{CaCO}}_3\underset{ }{\overset{ }{\to }}\text{CaO}+{\text{CO}}_2$

100gm        44gm
25gm                ?
Mass of ${\text{CO}}_{\text{2}}$  liberated from 100% pure 

$=\frac{25\times 44}{100}=11\mathrm{grams}$