$\mathrm{MgCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$

Let the weight of $\mathrm{CaCO}_3$ be $\mathrm{x} \mathrm{gm}$

$\therefore$weight of $\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{gm}$

Moles of $\mathrm{CaCO}_3$ decomposed $=$ moles of $\mathrm{CaO}$ formed

$\frac{\mathrm{x}}{100}=$ moles of $\mathrm{CaO}$ formed

$\therefore$ weight of $\mathrm{CaO}$ formed $=\frac{\mathrm{x}}{100} \times 56$

Moles of $\mathrm{MgCO}_3$ decomposed $=$ moles of $\mathrm{MgO}$ formed

$\frac{(2.21-x)}{84}=\text { moles of } \mathrm{MgO} \text { formed }$

$\therefore \text { weight of } \mathrm{MgO} \text { formed }=\frac{2.21-\mathrm{x}}{84} \times 40$

$\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152$

$\therefore \mathrm{x}=1.1886 \mathrm{~g}=\text { weight of } \mathrm{CaCO}_3$

& weight of $\mathrm{MgCO}_3=1.0214 \mathrm{~g}$