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Calcium carbonate decomposes on heating according to the equation;
${ CaCO }_{ 3 }{ (s)\rightarrow CaO(s)+ }{ CO }_{ 2 }{ (g) }$
At STP, the volume of carbon dioxide obtained by thermal decomposition of ${ 50g }$ of calcium carbonate will be:
(A) ${ 22.4litre }$
(B) ${ 44 litre }$
(C) ${ 11.2 litre }$
(D) ${ 1 litre }$

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Last updated date: 17th Apr 2024
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Answer
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Hint: Decomposition reaction can be defined as the splitting of a chemical compound or molecule into two or more molecules. These reactions are mostly endothermic. When a decomposition reaction occurs when the heat is provided to it, we say it as a thermal decomposition reaction.

Complete step by step solution:
> The chemical reaction takes place is:
${ CaCO }_{ 3 }{ (s)\rightarrow CaO(s)+ }{ CO }_{ 2 }{ (g) }$
As we can see here that 1 mol of calcium carbonate gives 1 mol of carbon dioxide.
> Molar mass of ${ CaCO }_{ 3 }$ = ${ 40+12+16\times 3 }$ = ${ 100g }$
So, according to the balanced reaction, ${ 100g }$ of ${ CaCO }_{ 3 }$ produces ${ 100g }$ of carbon dioxide.
As we know,
At STP, ${ 1 }$ mol of any gas = ${ 22.4L }$
Number of moles = ${ Mass\div molar\quad mass }$
= ${ 50\div 100 }$ = ${ 0.5mol }$
> Therefore, ${ 0.5mol }$ of ${ CaCO }_{ 3 }$ will produce = ${ 0.5\times 22.4L }$ = ${ 11.2L }$
Hence, at STP, ${ 11.2L }$ of carbon dioxide is obtained by thermal decomposition of ${ 50g }$ of calcium carbonate.

The correct option is C.

> ${ 1 }$ mole contains the same number of particles as there are in ${ 12g }$ of carbon-12 atoms. This number is called Avogadro’s number and is equal to ${ 6.023\times 10 }^{ 23 }$ particles.
${ 1 }$ mole of a gas occupies a volume of ${ 22.4 }$ litres at S.T.P.

Note: The possibility to make a mistake is that here ${ 1 mol }$ of calcium carbonate produces
${ 1 mol }$ of carbon dioxide not twice or half. Secondly, you must know the volume of gases at STP.