a
\(\mathrm{CaCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

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

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 \(\mathrm{MgCO}_3=1.0214 \mathrm{~g}\)