In an experiment, $1.0 \mathrm{g}$ of lime stone$\left({\mathrm{CaCO}}_3\right)$ gave $224 {\mathrm{cm}}^3$ of ${\mathrm{CO}}_2$at $\mathrm{NTP}$ and $0.56 \mathrm{g}$ of calcium oxide. These data illustrate the law of: 

${\mathrm{CaCO}}_3⟶\mathrm{CaO}+{\mathrm{CO}}_2\text{ gas }$

                 $0.56\mathrm{g}22400{\mathrm{cm}}^3$

wt. of ${\mathrm{CaCO}}_3=1.0\mathrm{g}$                 …(1)

wt. of $\mathrm{CaO}=0.56\mathrm{g}$

$22400{\mathrm{cm}}^3{\mathrm{CO}}_2\text{ weighs }=\text{ g. mol. wt. of }{\mathrm{CO}}_2$

                   $=12+(2\times 16)=44\mathrm{g}$

$\therefore 224{\mathrm{cm}}^3{\mathrm{CO}}_2\text{ weighs }=\frac{44\mathrm{g}}{22400}\times 224=0.44\mathrm{g}\text{. }$

$\therefore$wt. of products $=0.56+0.44=1.0\mathrm{g}$

wt. of reactant $\left({\mathrm{CaCO}}_3\right)=1.0\mathrm{g}$

Since the weight of reactants is equal to that of products, the law of conservation of mass is obeyed. So, the correct answer is (d).