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 $\mathrm{g} \mathrm{~mol}^{-1} \mathrm{CaCO}_3: 100, \mathrm{MgCO}_3: 84$)
Solution
<p>$$\begin{aligned}
& \mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \\
& \mathrm{MgCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})
\end{aligned}$$</p>
<p>Let the weight of $\mathrm{CaCO}_3$ be $\mathrm{x}$ gm</p>
<p>$\therefore$ weight of $\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{~gm}$</p>
<p>Moles of $\mathrm{CaCO}_3$ decomposed $=$ moles of $\mathrm{CaO}$ formed</p>
<p>$\frac{\mathrm{x}}{100}=$ moles of $\mathrm{CaO}$ formed</p>
<p>$\therefore$ weight of $\mathrm{CaO}$ formed $=\frac{\mathrm{x}}{100} \times 56$</p>
<p>Moles of $\mathrm{MgCO}_3$ decomposed $=$ moles of $\mathrm{MgO}$ formed</p>
<p>$\frac{(2.21-x)}{84}=$ moles of $\mathrm{MgO}$ formed</p>
<p>$\therefore$ weight of $\mathrm{MgO}$ formed $=\frac{2.21-\mathrm{x}}{84} \times 40$</p>
<p>$$\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152$$</p>
<p>$\therefore \mathrm{x}=1.1886 \mathrm{~g}=$ weight of $\mathrm{CaCO}_3$</p>
<p>& weight of $\mathrm{MgCO}_3=1.0214 \mathrm{~g}$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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