Question

What will be the volume of $C{O_2}$ at NTP obtained on heating $10grams$ of ($90\%$ pure) limestone?
$i)22.4{\text{ litres}}$
$ii)2.016{\text{ litres}}$
$iii)2.24{\text{ litres}}$
$iv)20.16{\text{ litres}}$

Answer 1

In order to solve this question, to find about the volume that will be produced, we need to have the understanding of the mole concept and the balancing of the chemical reaction. Also, there is an additional condition that $90\%$ pure limestone is given. Hence, that also plays a role in deciding the volume of the $C{O_2}$ that will be released in the process.

Complete answer:
In the given question, the limestone, i.e. $CaC{O_3}$ will be burnt to produce calcium oxide and carbon dioxide gas. Hence, in order to solve this question, we need to first understand and be able to write the complete and balanced chemical equation of the following reaction.
The reaction involved in the burning of limestone is: $CaC{O_3} \to CaO + C{O_2}$
Here, we can see that$1.0$mole of limestone gives off 1 mole i.e. $22.4{\text{ litres}}$ of $C{O_2}$at NTP.
Since $10grams$ of limestone is burnt, we need to convert this into moles.
Molar mass of limestone$= 40 + 12 + 48 = 100grams$(adding the individual atomic masses of calcium, carbon and three oxygen atoms)
Thus, number of moles in $10grams$of limestone$= \dfrac{{10}}{{100}} = 0.1moles$
Hence, with $0.1moles$, the moles of carbon dioxide released$= 0.1moles$
$1.0$ mole of $C{O_2}$ is equal to $22.4{\text{ litres}}$of$C{O_2}$at NTP. Thus, $0.1moles$ of $C{O_2}$ at NTP will be equal to :
$0.1 \times 22.4 = 2.24l$.
Now, since limestone is only $90\%$ pure , the actual volume of gas released will be $90\%$of the theoretically expected value.
Thus, $90\%$of $2.24l = 2.016litres$

Hence, the correct option is $(ii)$.

Note:
The general mistake attempted in these types of questions is where students omit the given detail of percentage of purity of the given entity, here, limestone. We need to take that into consideration whenever solving the question. Also , make sure that the formed chemical reaction is always balanced, otherwise the calculation of the number of moles will go wrong and which eventually will turn to the wrong calculation of the amount of mass required to be found in the question.