Question

What volume of oxygen gas \[\left( {{\text{O}}_{2}} \right)\] measured at \[0{}^\circ \text{C}\] and 1 atm, is needed to burn completely 1 L of propane gas \[\left( {{\text{C}}_{3}}{{\text{H}}_{8}} \right)\] measured under the same conditions?
A ) 5 L
B ) 10 L
C ) 7 L
D ) 6 L

Answer 1

Hint: Apply the Gay-Lussac's law of combining volumes.According to the Gay-Lussac's law of combining volumes, when two gases react, and if the temperature and pressure of two reacting gases are same, then volumes of two reacting gases have a simple whole number ratio.

Complete answer:
First write a balanced chemical equation for the combustion of propane. Then balance this equation. Using reaction stoichiometry, determine the ratio of number of moles of oxygen to number of moles of propane. Then correlate this ratio of number of moles to the ratio of volumes.
In presence of atmospheric oxygen, complete combustion of propane gas gives a mixture of carbon dioxide and water. Write an unbalanced chemical equation for the combustion process.

\[{{\text{C}}_{3}}{{\text{H}}_{8}}\text{ + }{{\text{O}}_{2}}\text{ }\to \text{ C}{{\text{O}}_{2}}\text{ + }{{\text{H}}_{2}}\text{O}\]

On the reactant side, three carbon atoms are present in propane molecule. On the products side only one carbon atom is present. To balance the number of carbon atoms, add coefficient three to carbon dioxide gas.

\[{{\text{C}}_{3}}{{\text{H}}_{8}}\text{ + }{{\text{O}}_{2}}\text{ }\to \text{ 3 C}{{\text{O}}_{2}}\text{ + }{{\text{H}}_{2}}\text{O}\]
On the reactant side, eight hydrogen atoms are present in the propane molecule. On the products side only two hydrogen atoms are present. To balance the number of hydrogen atoms, add coefficient four to water.

\[{{\text{C}}_{3}}{{\text{H}}_{8}}\text{ + }{{\text{O}}_{2}}\text{ }\to \text{ 3 C}{{\text{O}}_{2}}\text{ + 4 }{{\text{H}}_{2}}\text{O}\]

On the reactant side, two oxygen atoms are present in the dioxygen molecule. On the products side ten oxygen atoms are present. To balance the number of oxygen atoms, add coefficient five to dioxygen on the reactants side.
\[{{\text{C}}_{3}}{{\text{H}}_{8}}\text{ + 5 }{{\text{O}}_{2}}\text{ }\to \text{ 3 C}{{\text{O}}_{2}}\text{ + 4 }{{\text{H}}_{2}}\text{O}\]

This is a balanced chemical equation for the combustion of propane.
As per the reaction stoichiometry, one mole of propane reacts with five moles of oxygen.
The ratio of the volume of the oxygen to the volume of propane, measured under the same conditions of temperature and pressure, is equal to the ratio of the number of moles from the reaction stoichiometry. This ratio is 5:1. Thus, under identical conditions of temperature and pressure, 1 L of propane gas will react with 5 L of dioxygen gas.

Hence, the option A ) 5 L is the correct option.

Note: According to the Gay-Lussac's law of combining volumes, when two gases react, and if the temperature and pressure of two reacting gases are same, then volumes of two reacting gases have a simple whole number ratio.