QJEEM24Ch45c1: JEE Main 2024 Chemistry - Class 11 JEEM24Ch45c1 Solution: A sample of and weighed is ignited to constant weight of . The…

$\begin{aligned}& \mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \\& \mathrm{MgCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})\end{aligned}$

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 $\mathrm{MgCO}_3=1.0214 \mathrm{~g}$