Chapter 5, Problem 159CP

Interpretation Introduction

Interpretation: The formula of hydrocarbon in the given reaction of combustion of this hydrocarbon with ${\text{O}}_2$ at $\text{0}\text{.959}\text{atm}$ and $\text{298}\text{K}$ is needed to be identified, if the combustion producing a mixture of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ having a density of $\text{1}\text{.391}\text{g/L}$ and occupies a volume of 4 times larger than hydrocarbon.

Concept introduction:

• Number of moles of a substance,

• According to ideal gas equation,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

• Mass of a substance from its density and volume is,

$\text{Mass}\text{=}\text{Density}\times \text{Volume}$

Mass of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{taken}\text{Mass}$

• Balance equation of a reaction is written according to law of conservation of mass.
• Mole ratio between the reactants of a reaction is depends upon the coefficients of reactants in a balanced chemical equation.

Answer to Problem 159CP

Answer

The formula of hydrocarbon in the given reaction is ${\text{C}}_2{\text{H}}_6$.

Explanation of Solution

Explanation

• To find: the balanced chemical equation of the given reaction of combustion of hydrocarbon compound.

${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}{\text{O}}_{\text{2}}\to \text{x}{\text{CO}}_{\text{2}}\text{+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O}$

Let’s take the hydrocarbon as ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$.

The combustion of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ in ${\text{O}}_2$ is producing ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ gases.

Therefore,

The chemical equation of the reaction given is,

${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equation of the given reaction is,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen, hydrogen atoms and then oxygen atoms.

Therefore, the balanced chemical equation of the given reaction of combustion of hydrocarbon compound will be,

${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}{\text{O}}_{\text{2}}\to \text{x}{\text{CO}}_{\text{2}}\text{+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O}$

Hence,

The balanced chemical equation for the given reaction of combustion of hydrocarbon compound is, ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}{\text{O}}_{\text{2}}\to \text{x}{\text{CO}}_{\text{2}}\text{+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O}$.

The volume of hydrocarbon is taken as 1 L.

To find: the number of moles of reacted ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ in the given reaction.

The number of moles of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ is $\text{0}\text{.0392}\text{mol}$.

The pressure of reacted ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ in the given reaction is given as $\text{0}\text{.959}\text{atm}$.

The volume of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ is $\text{1}\text{L}$.

The temperature is given as $\text{298}\text{K}$.

According to ideal gas equation is,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Therefore, the number of moles of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$,

${\text{n}}_{{\text{C}}_{\text{x}}{\text{H}}_{\text{y}}}\text{=}\frac{\text{0}\text{.959}\text{atm}\text{×}\text{1}\text{L}}{\text{0}\text{.08206}\text{L}\text{atm/Kmol}\text{×}\text{298}\text{K}}\text{=}\text{0}\text{.0392}\text{mol}$

Hence, the number of moles of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ is in given reaction is $\text{0}\text{.0392}\text{mol}$.

The volume of mixture of products is taken as 4 L (since the volume of hydrocarbon is taken as 1 L).

To find: the number of moles and mass of produced mixture of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ having a density of $\text{1}\text{.391}\text{g/L}$ in the given reaction.

The number of moles of produced mixture is $\text{0}\text{.196}\text{mol}$.

The mass of produced mixture is $\text{5}\text{.564}\text{g}$.

The pressure of produced mixture in the given reaction is given as $\text{1}\text{.51}\text{atm}$.

The volume of produced mixture is $4\text{L}$.

The temperature is given as $\text{375}\text{K}$.

According to ideal gas equation is,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Therefore, the number of moles of produced mixture is,

${\text{n}}_{\text{mixture}}\text{=}\frac{\text{1}\text{.51}\text{atm}\text{×}4\text{L}}{\text{0}\text{.08206}\text{L}\text{atm/Kmol}\text{×}\text{375}\text{K}}\text{=}\text{0}\text{.196}\text{mol}$

Hence, the number of moles of produced mixture is in given reaction is $\text{0}\text{.196}\text{mol}$.

The density of produced mixture is given as $\text{1}\text{.391}\text{g/L}$

Mass of a substance from its density and volume is,

$\text{Mass}\text{=}\text{Density}\times \text{Volume}$

Therefore,

The mass of produced mixture is,

$\text{Mass}\text{=}\text{1}\text{.391}\text{g/L}\times \text{4}\text{L}\text{=}\text{5}\text{.564}\text{g}$

To find: the number of moles of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ in the produced mixture of given reaction.

The number of moles of ${\text{CO}}_{\text{2}}$ reacted is $\text{2}\times \text{0}\text{.0392}\text{mol}$.

The number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced is $\text{4}\times \text{0}\text{.0392}\text{mol}$

• The balanced chemical equation for the given reaction of combustion of hydrocarbon compound is, ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{+}{\text{O}}_{\text{2}}\to \text{x}{\text{CO}}_{\text{2}}\text{+}\frac{\text{y}}{\text{2}}{\text{H}}_{\text{2}}\text{O}$.

Here, the 1 mole of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ is producing $\text{x}\text{mol}{\text{CO}}_{\text{2}}\text{and}\frac{\text{y}}{\text{2}}\text{mol}{\text{H}}_{\text{2}}\text{O}$.

The number of moles of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ is in given reaction is calculated as $\text{0}\text{.0392}\text{mol}$.

Therefore,

$\text{0}\text{.0392}\text{mol}$ of ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$ is producing $\text{(0}\text{.0392)}\text{x}\text{mol}{\text{CO}}_{\text{2}}\text{and}(\text{0}\text{.0392)}\frac{\text{y}}{\text{2}}\text{mol}{\text{H}}_{\text{2}}\text{O}$.

The number of moles and mass of produced mixture of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is calculated as $\text{0}\text{.196}\text{mol}$.

That means,

$\text{(0}\text{.0392)}\text{x}\text{mol}+(\text{0}\text{.0392)}\frac{\text{y}}{\text{2}}\text{mol}=\text{0}\text{.196}\text{mol}\text{-Equation\_1}$

• The mass of produced mixture of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is calculated as $\text{5}\text{.564}\text{g}$.

The molecular mass of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ are 44g and 18g respectively.

Mass of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{taken}\text{Mass}$

Therefore,

$\text{(0}\text{.0392x)}(44)+(\text{0}\text{.0392}\frac{\text{y}}{\text{2}})(18)=\text{5}\text{.564}\text{g}\text{-Equation\_2}$

To solve the both equations,

Multiply the equation_1 with (44) and subtract with equation­_2.

That is,

$\begin{array}{l}\left[\text{(0}\text{.0392}\text{x)}(\text{44)}+(\text{0}\text{.0392)}\frac{\text{y}}{\text{2}}\text{(44)}=44\times \text{0}\text{.196}\right]-\\ \left[\text{(0}\text{.0392x)}(44)+(\text{0}\text{.0392}\frac{\text{y}}{\text{2}})(18)=\text{5}\text{.564}\text{g}\right]\\ =(\text{0}\text{.0392}\frac{\text{y}}{\text{2}})(26)=8.624-5.564\\ \text{0}\text{.0392}\frac{\text{y}}{\text{2}}=\frac{3}{26}\\ \text{y}=\frac{3\times \text{2}}{26\times \text{0}\text{.0392}}=6\\ \end{array}$

The value of y is found that 6, so the value of x will be 2 (since $\text{(0}\text{.0392)}\text{x}\text{mol}+(\text{0}\text{.0392)}\frac{\text{y}}{\text{2}}\text{mol}=\text{0}\text{.196}\text{mol}\text{-Equation\_1}$).

To determine: the formula of hydrocarbon ( ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$) in the given reaction of combustion of this hydrocarbon with ${\text{O}}_2$ at $\text{0}\text{.959}\text{atm}$.

The formula of hydrocarbon in the given reaction is ${\text{C}}_2{\text{H}}_6$.

The general formula of the hydrocarbon is ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$.

The values of x and y are calculated as 2 and 6.

Therefore, the formula of hydrocarbon in the given reaction is ${\text{C}}_2{\text{H}}_6$.

Conclusion

Conclusion

The formula of the reactant hydrocarbon of given reaction is identified. The formula of hydrocarbon ( ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}$) contains $\text{C}$ element and $\text{H}$ element, so it is depend upon the number of moles of produced ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ in the given reaction. The number of moles of these molecules is calculated from their given data’s.