Question

Calcium carbonate decomposes completely, on heating, into lime $({\text{CaO}})$ and carbon dioxide $(C{O_2})$. $1{\text{ kg}}$ of calcium carbonate is completely decomposed by heat, when $560{\text{ g}}$ of lime is obtained. How much quantity of carbon dioxide in grams, moles, and liters at ${\text{NTP}}$ is produced in the process?

Answer 1

The reaction given in the question involves the decomposition reaction which means we can use the law of conservation of mass to find out the answer. One can draw a reaction scheme to better analyze and understand the reaction and compare the number of moles which can lead to obtaining value of carbon dioxide produced in the process.

Complete step by step answer:

1. First of all the given reaction is the type of decomposition reaction which means one compound is chemically decomposed into two compounds which means according to the law of conservation of mass we can say that,
   ${\text{Mass of calcium carbonate = Mass of lime (CaO) + Mass of carbon dioxide (C}}{{\text{O}}_2})$
2. So we can calculate the value of the mass of carbon dioxide by using the given values of calcium carbonate and lime as follows,
   ${\text{Mass of carbon dioxide (C}}{{\text{O}}_2}){\text{ = Mass of calcium carbonate - Mass of lime (CaO) }}$
   ${\text{Mass of carbon dioxide (C}}{{\text{O}}_2}){\text{ = 1000 - 560}}$
   ${\text{Mass of carbon dioxide (C}}{{\text{O}}_2}){\text{ = 440 g}}$
   Therefore, The mass of carbon dioxide present is ${\text{440 g}}$.
3. Now that we have got the value of the mass of carbon dioxide we can calculate its number of moles which are present in the ${\text{440 g}}$which has molar mass as follows,
   ${\text{C}}{{\text{O}}_2}{\text{ = 12 + (2}} \times {\text{16) = 12 + 32 = 44 g}}$
   Now, let's put this value in finding out the number of moles formula,
   ${\text{Number of moles of C}}{{\text{O}}_2}{\text{ in 440 g = }}\dfrac{{{\text{Total mass}}}}{{{\text{Molar mass}}}} = \dfrac{{440}}{{44}} = 10{\text{ moles}}$
   Therefore, the number of moles carbon dioxide present is ${\text{10 moles}}$.
4. Now let us find out the value in liters as ${\text{1 mole}}$ of carbon dioxide occupies the $22 \cdot 4{\text{ litre}}$ of volume at ${\text{NTP}}$ so we can find out the value for the ${\text{10 moles}}$ which is,
   $= 22 \cdot 4{\text{ }} \times {\text{ 10 = 224 litre}}$
   Therefore, the ${\text{10 moles}}$ of carbon dioxide occupies the volume of $224{\text{ litre}}$.
5. Hence, the quantity of carbon dioxide produced is ${\text{440 g}}$, ${\text{10 moles}}$ and $224{\text{ litre}}$ at ${\text{NTP}}$ in the process.

Note:
${\text{NTP}}$ stands for normal temperature and pressure. The law of conservation of mass described as mass is neither created nor destroyed in a chemical reaction which means the mass of all the reactants and mass of all the products is always the same. Once one finds out the mass of a chemical compound then the number of moles and its volume can be calculated hence it is important to calculate the mass of the compound first.