Chapter 9, Problem 50E

Interpretation Introduction

Interpretation:

The number of moles of ${\text{CO}}_2$, the volume occupied by the number moles of ${\text{CO}}_2$ gas, and the number of years required to double the amount of ${\text{CO}}_2$ are to be calculated.

Concept Introduction:

Mole is the unit for calculating large quantities of small units, such as atoms and molecules.

Carbon dioxide is a colorless and odorless gas.

A small amount of carbon dioxide is essential for sustaining life on the earth.

Answer to Problem 50E

Solution:

a)

The number of moles of ${\text{CO}}_2$ is $2.4\times {10}^{14}\text{mole}$.

b)

The volume of ${\text{CO}}_2$ produced is $\text{6}\text{.3}\times {10}^{15}\text{L}$.

c)

The number of years required to double the amount of ${\text{CO}}_2$ is $240\text{years}$.

Explanation of Solution

a) Number of moles of ${\text{CO}}_2$ are produced by the combustion of $4\times {10}^{12}\text{Kg}$ of ${\text{C}}_8{\text{H}}_{18}$.

The amount of ${\text{CO}}_2$ present in the atmosphere is $0.04\%$ and the world uses $4\times {10}^{12}\text{Kg}$ of petroleum per year.

The reaction is as follows:

${\text{2C}}_8{\text{H}}_{18}\left(l\right)+25{\text{O}}_2\left(g\right)\to 16{\text{CO}}_2\left(g\right)+18{\text{H}}_2\text{O}\left(g\right)$.

The equation for the reaction of ${\text{C}}_8{\text{H}}_{18}$ with ${\text{O}}_2$ is as follows:

${\text{2C}}_8{\text{H}}_{18}\left(l\right)+25{\text{O}}_2\left(g\right)\to 16{\text{CO}}_2\left(g\right)+18{\text{H}}_2\text{O}\left(g\right)$

The mass of ${\text{C}}_8{\text{H}}_{18}$ is converted into grams as follows:

$\begin{array}{l}1\text{Kg}=1000g\\ \text{The}\text{mass}\text{of}\text{the}{\text{C}}_8{\text{H}}_{18}\text{in}\text{grams}=4.0\times {10}^{12}\times 1000g\\ =4.0\times {10}^{15}g\end{array}$

The moles of ${\text{C}}_8{\text{H}}_{18}$ are calculated using the expression as follows:

${\text{Moles of C}}_8{\text{H}}_{18}=\frac{\text{Given}\text{Mass}}{\text{Molecular}\text{Mass}}$.

Substitute the value of given mass and molecular mass in the above expression.

$\begin{array}{l}{\text{Moles of C}}_8{\text{H}}_{18}=\frac{4.0\times {10}^{15}\text{g}{\text{C}}_8{\text{H}}_{18}}{114.2g{\text{C}}_8{\text{H}}_{18}}\\ =3.5\times {10}^{13}\text{mole}\end{array}$.

From the above equation, 2 moles of ${\text{C}}_8{\text{H}}_{18}$ will produce 16 moles of ${\text{CO}}_2$. Therefore, moles of ${\text{CO}}_2$ will be $\frac{2}{16}$ of the moles of ${\text{C}}_8{\text{H}}_{18}$.

$\begin{array}{l}{\text{Moles of CO}}_2=\frac{2}{16}\times 3.5\times {10}^{13}\text{mole}\\ =2.8\times {10}^{14}\text{mole}\end{array}$

Hence, the number of moles ${\text{CO}}_2$ is $2.4\times {10}^{14}\text{mole}$.

b) The volume occupied by number of moles of ${\text{CO}}_2$ gas that just produced.

1 mole of gas occupies 22.4L.

The volume of ${\text{CO}}_2$ produced is calculated by the expression as follows:

${\text{Volume of CO}}_2={\text{Moles of CO}}_2\times 22.4\text{L}$.

Substitute the value of moles of ${\text{CO}}_2$ in the above expression.

$\begin{array}{l}{\text{Volume of CO}}_2=2.8\times {10}^{14}\text{mole}\times 22.4\text{L}\\ \text{=}\text{6}\text{.3}\times {10}^{15}\text{L}\end{array}$.

Hence, the volume of ${\text{CO}}_2$ produced is $\text{6}\text{.3}\times {10}^{15}\text{L}$.

c) The number of years it will take to double the amount of ${\text{CO}}_2$ currently present in the atmosphere.

The volume of ${\text{CO}}_2$ present in the atmosphere is $1.5\times {10}^{18}\text{L}$.

The volume of ${\text{CO}}_2$ present in the atmosphere is $1.5\times {10}^{18}\text{L}$. All ${\text{CO}}_2$ is produced by the combustion of petroleum. The number of years it will take to double the amount of ${\text{CO}}_2$ is calculated as follows:

$\text{Number}\text{of}\text{year}=\frac{\text{volume}\text{of}\text{present}{\text{CO}}_2}{\text{volume}\text{of}{\text{CO}}_2\text{produced}}$.

Substitute the volume of ${\text{CO}}_2$ present and the volume of ${\text{CO}}_2$ produced in the above expression.

$\begin{array}{l}\text{Number}\text{of}\text{year}=\frac{1.5\times {10}^{18}\text{L}}{6.3\times {10}^{15}\text{L}}\\ =240\text{years}\end{array}$

Hence, the number of years it will take to double the amount of ${\text{CO}}_2$ is $240\text{years}$.