Chapter 3, Problem 3.147QP

A mixture of methane (CH₄) and ethane (C₂H₆) of mass 13.43 g is completely burned in oxygen. If the total mass of CO₂ and H₂O produced is 64.84 g, calculate the fraction of CH₄ in the mixture.

Interpretation Introduction

Interpretation:

The fraction of ${\text{CH}}_{\text{4}}$ in the mixture of the given reaction has to be determined when $\text{13}\text{.43g}$ of mixture produces ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ in a total mass of $\text{64}\text{.84g}$.

Concept introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Mole ratio between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.

Answer to Problem 3.147QP

The fraction of ${\text{CH}}_{\text{4}}$ in the mixture is $\text{0}\text{.387}$.

Explanation of Solution

In given reaction,

Mixture of ${\text{CH}}_{\text{4}}$ and ${\text{C}}_2{\text{H}}_6$ are burned in ${\text{O}}_2$.

Therefore,

The chemical equation for this reaction is,

${\text{CH}}_4\text{+}{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$

${\text{C}}_2{\text{H}}_6\text{+}{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Therefore,

The total number of each atoms in the reactant side should equal to the total number of each atoms in the product side.

So, in order to balance a chemical equation, the coefficients of compounds or atoms are needed to be changed in such a way that total number of each atoms in the reactant side and the total number of each atoms in the product side is to become equal.

Hence,

The balanced equations for the given reactions are,

${\text{CH}}_4\text{+}2{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$

${\text{2C}}_2{\text{H}}_6\text{+}7{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$

Assumes mass of ${\text{CH}}_{\text{4}}$is ‘x’ and mass of ${\text{C}}_2{\text{H}}_6$is ‘13.43-x’.

The mass of Mixture of ${\text{CH}}_{\text{4}}$ and ${\text{C}}_2{\text{H}}_6$ is given as $\text{13}\text{.43g}$.

Let’s take mass of ${\text{CH}}_{\text{4}}$ is x.

So, the number of moles of ${\text{CH}}_{\text{4}}$ in the reaction is,

$\text{x g}{\text{CH}}_{\text{4}}\text{×}\frac{\text{1mol}{\text{CH}}_{\text{4}}}{\text{16}\text{.04g}{\text{CH}}_{\text{4}}}\text{=}\frac{\text{x}}{\text{16}\text{.04}}\text{mol}$.

The balanced chemical equation of the reaction is,

${\text{CH}}_4\text{+}2{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$

The mole ratio between ${\text{CH}}_{\text{4}}$ and ${\text{CO}}_{\text{2}}$ in the reaction is 1:1.

The mole ratio between ${\text{CH}}_{\text{4}}$ and ${\text{H}}_{\text{2}}\text{O}$ in the reaction is 1:2.

So, the number of moles of ${\text{CO}}_{\text{2}}$ produced in the reaction is $\frac{\text{x}}{\text{16}\text{.04}}\text{mol}$ and the number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced in the reaction is $\frac{\text{2x}}{\text{16}\text{.04}}\text{mol}\text{=}\frac{\text{x}}{\text{8}\text{.02}}\text{mol}$.

Then,

The mass of ${\text{CO}}_{\text{2}}$ is,

$\frac{\text{x}}{\text{16}\text{.04}}\text{mol}{\text{CO}}_{\text{2}}\text{×}\text{44}\text{.01g}{\text{CO}}_{\text{2}}\text{=}\text{2}\text{.744x}\text{g}{\text{CO}}_{\text{2}}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is,

$\frac{\text{x}}{\text{8}\text{.02}}\text{mol}{\text{H}}_{\text{2}}\text{O}\text{×}\text{18}\text{.02g}{\text{H}}_{\text{2}}\text{O}\text{=}\text{2}\text{.247x}\text{g}{\text{H}}_{\text{2}}\text{O}$.

The mass of Mixture of ${\text{CH}}_{\text{4}}$ and ${\text{C}}_2{\text{H}}_6$ is given as $\text{13}\text{.43g}$.

Let’s take mass of ${\text{C}}_2{\text{H}}_6$ is 13.43-x.

So, the number of moles of ${\text{C}}_2{\text{H}}_6$ in the reaction is,

$\text{(13}\text{.43-x) g}{\text{C}}_2{\text{H}}_6\text{×}\frac{\text{1mol}{\text{C}}_2{\text{H}}_6}{\text{30}\text{.07g}{\text{C}}_2{\text{H}}_6}\text{=}\frac{\text{13}\text{.43-x}}{\text{30}\text{.07}}\text{mol}{\text{C}}_2{\text{H}}_6$

The balanced chemical equation of the reaction is,

${\text{2C}}_2{\text{H}}_6\text{+}7{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$

The mole ratio between ${\text{C}}_2{\text{H}}_6$ and ${\text{CO}}_{\text{2}}$ in the reaction is 1:2.

The mole ratio between ${\text{C}}_2{\text{H}}_6$ and ${\text{H}}_{\text{2}}\text{O}$ in the reaction is 1:3.

So, the number of moles of ${\text{CO}}_{\text{2}}$ produced in the reaction is $\frac{\text{2(13}\text{.43-x)}}{\text{30}\text{.07}}\text{mol}\text{=}\frac{\text{13}\text{.43-x}}{\text{15}\text{.035}}\text{mol}$ and the number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced in the reaction is $\frac{\text{3(13}\text{.43-x)}}{\text{30}\text{.07}}\text{mol}\text{=}\frac{\text{13}\text{.43-x}}{\text{10}\text{.02}}\text{mol}$.

Then,

The mass of ${\text{CO}}_{\text{2}}$ is,

$\frac{\text{13}\text{.43-x}}{\text{15}\text{.035}}\text{mol}{\text{CO}}_{\text{2}}\text{×}\text{44}\text{.01g}{\text{CO}}_{\text{2}}\text{=}\text{2}\text{.927(13}\text{.43-x)g}{\text{CO}}_{\text{2}}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is,

$\frac{\text{13}\text{.43-x}}{\text{10}\text{.02}}\text{mol}{\text{H}}_{\text{2}}\text{O}\text{×}\text{18}\text{.02g}{\text{H}}_{\text{2}}\text{O}\text{=}\text{1}\text{.798(13}\text{.43-x)}\text{g}{\text{H}}_{\text{2}}\text{O}$.

The mass of ${\text{CO}}_{\text{2}}$ is produced by ${\text{CH}}_{\text{4}}$ is found as $\text{2}\text{.744x}\text{g}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is produced by ${\text{CH}}_{\text{4}}$ is found as $\text{2}\text{.247x}\text{g}$.

The mass of ${\text{CO}}_{\text{2}}$ is produced by ${\text{C}}_2{\text{H}}_6$ is found as $\text{2}\text{.927(13}\text{.43-x)}\text{g}$.

The mass of ${\text{H}}_{\text{2}}\text{O}$ is produced by ${\text{C}}_2{\text{H}}_6$ is found as $\text{1}\text{.798(13}\text{.43-x)}\text{g}$.

The total mass of ${\text{CO}}_{\text{2}}\text{and}{\text{H}}_{\text{2}}\text{O}$ produced by the mixture is given that $\text{64}\text{.84g}$.

Therefore,

$\begin{array}{l}\text{2}\text{.744x}\text{g}\text{+}\text{2}\text{.247x}\text{g}\text{+}\text{2}\text{.927(13}\text{.43-x)}\text{g}\text{+}\text{1}\text{.798(13}\text{.43-x)}\text{g}\text{=}\text{64}\text{.84g}\\ \text{0}\text{.266x}\text{g}\text{+}\text{63}\text{.4568g}\text{=}\text{64}\text{.84g}\\ \text{0}\text{.266x}\text{g}\text{=}\text{1}\text{.383g}\\ \text{x}\text{=}\text{5}\text{.2}\end{array}$

So, the mass of ${\text{CH}}_{\text{4}}$ is 5.2g.

Hence,

The fraction of ${\text{CH}}_{\text{4}}$ in the mixture is,

$\frac{\text{5}\text{.2g}}{\text{13}\text{.43g}}\text{=}\text{0}\text{.387}$.

Conclusion

The fraction of ${\text{CH}}_{\text{4}}$ in the mixture of the given reaction was determined as $\text{0}\text{.387}$