JEE Main 2024 — Some Basic Concepts Of Chemistry Question with Solution

$$\begin{array}{rl}& {\mathrm{CaCO}}_3(\mathrm{s})\stackrel{\Delta }{\to }\mathrm{CaO}(\mathrm{s})+{\mathrm{CO}}_2(\mathrm{g})\\ & {\mathrm{MgCO}}_3(\mathrm{s})\stackrel{\Delta }{\to }\mathrm{MgO}(\mathrm{s})+{\mathrm{CO}}_2(\mathrm{g})\end{array}$$

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

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

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

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

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

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