JEE MAIN - Chemistry (2024 - 31st January Evening Shift - No. 20)
A sample of $$\mathrm{CaCO}_3$$ and $$\mathrm{MgCO}_3$$ weighed $$2.21 \mathrm{~g}$$ is ignited to constant weight of $$1.152 \mathrm{~g}$$. The composition of mixture is :
(Given molar mass in $$\mathrm{g} \mathrm{~mol}^{-1} \mathrm{CaCO}_3: 100, \mathrm{MgCO}_3: 84$$)
Explanation
$$\begin{aligned} & \mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \\ & \mathrm{MgCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \end{aligned}$$
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 $$\mathrm{MgCO}_3=1.0214 \mathrm{~g}$$
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