Chapter 13.3, Problem 38PP

Interpretation Introduction

Interpretation: The volume of propane gas is to be calculated when it undergoes combustion with a definite volume of oxygen gas.

Concept introduction: The given reaction is a combustion reaction. Volume-volume stochiometry and Avogadro’s Law will help to solve the problem.

Answer to Problem 38PP

The volume of propane undergoing combustion is $6.8\text{L}$ by $34\text{L}$ oxygen.

Explanation of Solution

According to Avogadro’s Law, equal volumes of all gases under specific temperature and pressure contain an equal number of molecules.

The balanced chemical equation is given by:

  ${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 3\text{C}{\text{O}}_2+4{\text{H}}_2\text{O}$

The reaction shows that one molecule of methane combines with five molecules of oxygen in the balanced chemical equation.

Therefore,

$5\text{L}$ of oxygen reacts with $1\text{L}$ propane gas

$34\text{L}$ of oxygen reacts with $\begin{array}{l}=\frac{34}{5}\text{L}\end{array}$

           $\begin{array}{c}=6.8\text{L}\end{array}$

Conclusion

$6.8\text{L}$ of propane gas undergo combustion using $34\text{L}$ oxygen gas.