Question

Calcium carbonate decomposes on heating according to the equation;
${CaCO}_3(s)\to CaO(s)+{CO}_2(g)$
At STP, the volume of carbon dioxide obtained by thermal decomposition of $50g$ of calcium carbonate will be:
(A) $22.4litre$
(B) $44litre$
(C) $11.2litre$
(D) $1litre$

Answer 1

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:
${CaCO}_3(s)\to CaO(s)+{CO}_2(g)$
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 = $Mass÷molarmass$
= $50÷100$ = $0.5mol$
> 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 $1mol$ of calcium carbonate produces
$1mol$ of carbon dioxide not twice or half. Secondly, you must know the volume of gases at STP.