Question

When \[10g\] of \[90\% \] pure limestone is heated, the volume of \[C{O_2}\] (in liter) liberated at STP is:
A.\[22.4L\]
B.\[2.24L\]
C.\[20.16L\]
D.\[2.016L\]

Answer 1

Hint: Depending on the method of preparation, limestone may be granular, crystalline, clastic or dense. The limestone is prepared by using calcite and the pure type of limestone is known as chalk. Here, we measure the volume of liberated carbon dioxide at STP and it means the standard temperature and pressure. It is defined in the terms of zero degree Celsius at one atmosphere of pressure.

Complete answer:
The amount of liberated carbon dioxide is not equal to\[22.4L\]. At STP, one mole of gas is only occupied at \[22.4L\] and here it is not equal to one mole. Hence, option (A) is incorrect.
The volume of carbon dioxide liberated in a liter is not equal to \[2.24L\]. Hence, the option (B) is incorrect.
By heating of \[10g\] \[90\% \] pure limestone, there is a liberation of carbon dioxide but it is not equal to \[20.16L\]. Hence, option (C) is incorrect.
By heating of calcium carbonate, (limestone) there is a formation of calcium oxide with liberation of carbon dioxide. And the reaction can be written as,
\[CaC{O_3}\xrightarrow{\Delta }CaO + C{O_2}\]
Here, the ratio of calcium carbonate and carbon dioxide is equal to \[1:1\]
\[10g\] \[90\% \]Pure limestone =$9g$ \[CaC{O_3}\]
Molecular weight of calcium carbonate\[ = 100g/mol\]
Hence, Number of moles of \[CaC{O_3}\] is heated\[ = \dfrac{9}{{100}}moles\]
Therefore, for \[\dfrac{9}{{100}}moles\] calcium carbonate, it will liberate \[\dfrac{9}{{100}}moles\] of carbon dioxide. Thus, \[\dfrac{9}{{100}}moles\,CaC{O_3}:\,\dfrac{9}{{100}}moles\,\]liberated \[C{O_2}\]
One mole if carbon dioxide is equal to \[22.4L\] at STP
Hence, \[\dfrac{9}{{100}}moles\]of \[C{O_2}\]\[ = \_\_\,L\]at STP
Volume of \[C{O_2}\] liberated at STP\[ = \dfrac{9}{{100}}x22.4\]
\[ = 2.016L\]

Hence, the option (D) is correct.

Note:
By heating of \[10g\] \[90\% \] pure limestone, the volume of \[C{O_2}\] liberated at STP is equal to \[2.016L\]. When one mole of calcium carbonate is heated, there is a formation of one mole of calcium oxide with liberation of one mole of carbon dioxide. And the ratio of calcium carbonate and carbon dioxide is equal to \[1:1\]. And here, the liberated carbon dioxide is found in the term of STP.