Question

Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide. How much volume of $C{O_2}$ will be obtained at STP by thermal decomposition of 50 g of $CaC{O_3}$ ?
A.1L
B.11.2L
C.44L
D.22.4L

Answer 1

Hint: To solve this question, we shall first write a balanced equation of the reaction. We shall then find the amount of $CaC{O_3}$ reacting in moles, which will also give us the amount of $C{O_2}$ formed. From here, we can calculate the volume of gas from the ideal gas equation at STP.

Formula used: $n = \dfrac{w}{{{M_0}}}$ (Eq1)
where n is number of moles, w is the weight of the compound and ${M_0}$ is the molecular weight of the compound.
$PV = nRT$ (Eq2)
where P is pressure of the gas, V is volume of the gas, n is the number of moles of gas, R is a constant ($0.0821{\text{ }}atm{\text{ }}L{\text{ }}{K^{ - 1}}mo{l^{ - 1}}$) and T is the temperature.

Complete step by step answer:
From the question, we know that calcium carbonate on heating forms calcium oxide and carbon dioxide. The balanced equation of this reaction will be as follows:
$CaC{O_3} \to CaO + C{O_2}$ (Eq. 3)
Now, we find the moles of $CaC{O_3}$ present:
Let the number of moles of $CaC{O_3}$ be ${n_{CaC{O_3}}}$ and ${M_0}$ of $CaC{O_3}$ is 100 g/mol.
Putting the values in Eq. 1, we get:
${n_{CaC{O_3}}} = \dfrac{{50}}{{100}}$
$ \Rightarrow {n_{CaC{O_3}}} = 0.5$
So, from Eq. 3, using stoichiometry, we can conclude that the number of moles of $CaC{O_3}$ is equal to the number of moles of $C{O_2}$.
$ \Rightarrow {n_{C{O_2}}} = 0.5$
Now, we shall calculate the volume of $C{O_2}$ present using the ideal gas equation, i.e. Eq. 2:
At STP, P=1 atm and T=273 K.
Putting these values in Eq. 2, we get:
$ \Rightarrow V = \dfrac{{0.5 \times 0.0821 \times 273}}{1}$
$ \Rightarrow V = 11.2L$

$\therefore $ The correct option is option B, i.e. 11.2L. .

Note:
The volume of an ideal gas at STP can also be easily calculated from the simplified gas equation: $V = n \times 22.4$ . So, according to the question, we found that $n = 0.5$.
Substituting the value of n in $V = n \times 22.4$, we get:
$ \Rightarrow V = 0.5 \times 22.4$
$ \Rightarrow V = 11.2L$