Chapter 13.3, Problem 40PP

Interpretation Introduction

Interpretation:

Volume of oxygen gas is to be calculated when it is used to combust a definite volume of methane gas.

Concept introduction:

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

Answer to Problem 40PP

The volume of oxygen needed to combust $2.36L$ methane is $4.72L$

Explanation of Solution

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

The balanced chemical equation is given by:

${\mathrm{CH}}_4+2O_2\to C{O}_2+2H_{2}O$

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

Therefore volume of oxygen needed to combust $2.36L$ methane

$\begin{array}{l}=2\times 2.36\\ =4.72L\end{array}$

Conclusion

$4.72L$ Oxygen is required to combust $2.36L$ methane gas.