Chapter 5, Problem 5.62QP

Ethanol (C₂H₅OH) burns in air:

${\text{C}}_2{\text{H}}_5\text{OH}(l)+{\text{O}}_2(g)\stackrel{}{\to }{\text{CO}}_2(g)+{\text{H}}_2\text{O}(l)$

Balance the equation and determine the volume of air in liters at 35.0°C and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent O₂ by volume.

Interpretation Introduction

Interpretation:

The given equation has to be balanced and the volume of air in liters has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

  $\text{PV = nRT}$

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge are the similar for both the reactants and the products.

Answer to Problem 5.62QP

The balanced equation given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + 3O}}_{\text{2}}{}_{\text{(g)}}\to {\text{ 2CO}}_{\text{2}}\text{(g)}\text{ +}{\text{3H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The volume of air was found to be $\text{1}{\text{.71×10}}^{\text{3}}\text{L}$.

Explanation of Solution

Unbalanced equation is written below:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2}}{}_{\text{(g)}}\to {\text{ CO}}_{\text{2}}\text{(g)}\text{ +}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Carbon is balanced by multiplying 2 in product side

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + O}}_{\text{2}}{}_{\text{(g)}}\to {\text{ 2CO}}_{\text{2}}\text{(g)}\text{ +}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Next Oxygen is balance by multiplying 3 in reactant side

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + 3O}}_{\text{2}}{}_{\text{(g)}}\to {\text{ 2CO}}_{\text{2}}\text{(g)}\text{ +}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Finally Oxygen and Hydrogen is balanced by adding coefficient 3 before water in product side.

Hence, the balanced equation given as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}{\text{ + 3O}}_{\text{2}}{}_{\text{(g)}}\to {\text{ 2CO}}_{\text{2}}\text{(g)}\text{ +}{\text{3H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The moles of ${\text{O}}_{\text{2}}$:

The moles of ${\text{O}}_{\text{2}}$ is calculated by plugging in the values of the given molecular weight of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ and weight of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$.

  $\text{227}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH×}\frac{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{46}\text{.07}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{×}\frac{\text{3}\text{mol}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{=14}\text{.8}\text{mol}{\text{O}}_{\text{2}}$

The moles of ${\text{O}}_{\text{2}}$ was found to be $\text{14}\text{.8}\text{mol}{\text{O}}_{\text{2}}$

The volume of ${\text{O}}_{\text{2}}$using the ideal gas equation.

The volume of ${\text{O}}_{\text{2}}$ gas is calculated by plugging in the values of the given moles, temperature and pressure.

  $\begin{array}{c}{\text{ V}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{RT}}{\text{P}}\\ {\text{ V}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{(14}\text{.8}{\text{mol O}}_{\text{2}}\text{)}\left(\text{0}\text{.08206}\frac{\text{L}\cdot \text{atm}}{\text{K}\cdot \text{mol}}\right)\text{(318)}\text{K}}{\left(\text{793}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)}\\ \text{=}\text{3}\text{.60}\times {\text{10}}^{\text{2}}\text{L}\end{array}$

The volume of ${\text{O}}_{\text{2}}$ gas was found to be $\text{3}\text{.60}\times {\text{10}}^{\text{2}}\text{L}$.

In air 21.0 percent is ${\text{O}}_{\text{2}}$by volume

The volume of air is calculated by plugging in the values of the given volume of ${\text{O}}_{\text{2}}$ and percentage of air.

  $\begin{array}{c}{\text{ V}}_{\text{air}}{\text{=V}}_{{\text{O}}_{\text{2}}}\times \left(\frac{\text{100\%}\text{air}}{\text{21\%}{\text{O}}_{\text{2}}}\right)\\ \text{=(360}\text{L}{\text{O}}_{\text{2}}\text{)}\left(\frac{\text{100\%}\text{air}}{\text{21\%}{\text{O}}_{\text{2}}}\right)\\ \text{=}\text{1}{\text{.71×10}}^{\text{3}}\text{L}\text{air}\end{array}$

The volume of air was found to be $\text{1}{\text{.71×10}}^{\text{3}}\text{L}$.

Conclusion

The given equation was balanced and the volume of air in liters was calculated.