Question

How much volume of $C{O_2}$ at S.T.P is liberated by the combustion of 100 $c{m^3}$ of propane $({C_3}{H_8})$?
A. 100 $c{m^3}$
B. 200 $c{m^3}$
C. 300 $c{m^3}$
D. 400 $c{m^3}$

Answer 1

The elements in nature exist in one of the three states, solid, liquid or gas. The gases are the freest in their mobility while the solids are the most rigid. All these properties arise from the intermolecular forces of attraction that act between the molecules and atoms of the elements. All the reactions that take place are along with a specific ratio in which the elements react.

Complete step by step answer:
S.T.P is the acronym for the standard temperature and pressure. It defines the state in which the temperature and pressure are defined and the volume taken by any gas is found to be 22.4 L.
1 mole of all the gases at S.T.P is found to acquire a volume of 22.4L.
The chemical reaction that will take place in the above question can be stated as follows
${C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O$
Analysing the above reaction and using the stoichiometric coefficients we see that we get 3 times the carbon dioxide of the propane used. So for 1-mole propane used we get 3 moles of carbon dioxide.
The volume of 1-mole gas at STP is 22.4L
The moles of 100 $c{m^3}$ propane gas will be $\dfrac{{100c{m^3}}}{{22.4}}$
Thus 100 $c{m^3}$ of propene at STP will form
$\dfrac{{100c{m^3}}}{{22.4}} \times 3 \times 22.4L$
$= 300c{m^3}$ at STP of carbon dioxide.

so, the correct answer to the above question is C.

Note: Charles law is a gas law which is derived after experimentation. It describes how gases tend to increase in volume when they are heated. The law states that, when the pressure of the gas is held constant, the kelvin temperature and the volume of the gas will be directly proportional. This law can be mathematically stated as
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$
Where ${V_1}$ is the first volume and ${V_2}$ is the second volume. Similarly, ${T_1}$ is the first temperature and ${T_2}$ is the second temperature.