Chapter 21, Problem 21.59QE

(a)

Interpretation Introduction

Interpretation:

Amount of coal that is required to produce the energy as same as fission of $\text{1}\text{kg}$ uranium has to be calculated in metric tons.

Explanation of Solution

Given information:

Amount of energy produced from one kilogram of coal is $2.8\times {10}^4\text{kJ}$.  Amount of energy produced by fission of $\text{1}\text{mol}$ of ${}^{\text{235}}\text{U}$ is $1.9\times {10}^{10}\text{kJ}$.

Molar mass of ${}^{\text{235}}\text{U}$ is $\text{235}\text{.05}\text{g/mol}$.  Therefore, mass of one mole of ${}^{\text{235}}\text{U}$ is $\text{235}\text{.05}\text{g}$.

Energy released by one mole of ${}^{\text{235}}\text{U}$ is $1.9\times {10}^{10}\text{kJ}$.  This is equal to energy released by $\text{0}\text{.23505}\text{kg}$ of ${}^{\text{235}}\text{U}$.

Amount of energy released by one kilogram of uranium-235 is calculated as shown below.

    $\begin{array}{c}\text{Energy}\text{released}=\frac{1.9\times {10}^{10}\text{kJ}}{0.23505\text{kg}}\times 1\text{kg}\\ \text{=8}\text{.083}\times {\text{10}}^{\text{10}}\text{kJ}\\ \text{=8}\text{.1}\times {\text{10}}^{\text{10}}\text{kJ}\end{array}$

Therefore, one kilogram of uranium-235 releases $\text{8}\text{.1}\times {\text{10}}^{\text{10}}\text{kJ}$ of energy.

One kilogram of high-grade coal produces $2.8\times {10}^4\text{kJ}$ of energy.  Mass of coal required to produce $\text{8}\text{.1}\times {\text{10}}^{\text{10}}\text{kJ}$ of energy can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{of}\text{coal}=8.1\times {10}^{10}\text{kJ}\times \frac{1\text{kg}}{2.8\times {10}^4\text{kJ}}\\ =2.892\times {10}^6\text{kg}\\ \text{=2}\text{.9}\times {10}^6\text{kg}\end{array}$

Conversion of kilogram into metric ton can be done using the conversion factor as shown below.

    $\begin{array}{c}\text{2}\text{.9}\times {10}^6\text{kg}=\text{2}\text{.9}\times {10}^6\text{kg}\times \frac{\text{1}\text{metric}\text{ton}}{\text{1000}\text{kg}}\\ =\text{2}{\text{.9×10}}^{\text{3}}\text{metric}\text{ton}\end{array}$

Therefore, the amount of coal required to produce the same amount of energy as of one kilogram of uranium-235 is $\text{2}{\text{.9×10}}^{\text{3}}\text{metric}\text{ton}$.

(b)

Interpretation Introduction

Interpretation:

Metric tons of sulfur dioxide that is produced by burning $\text{2}{\text{.9×10}}^{\text{3}}\text{metric}\text{ton}$ of coal has to be calculed.

Explanation of Solution

In the problem statement it is given that coal contains $0.90\%$ mass sulfur.  This means $\text{0}\text{.9}\text{g}$ of sulfur is present in $\text{100}\text{g}$ of coal.  Total grams of sulfur present in $\text{2}{\text{.9×10}}^{\text{3}}\text{metric}\text{ton}$ of coal is calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{of}\text{Sulfur}=\text{2}{\text{.9×10}}^{\text{3}}\text{metric}\text{ton coal}\times \frac{\text{0}\text{.9}\text{g}\text{S}}{\text{100}\text{g}\text{coal}}\\ =\text{2}{\text{.9×10}}^{\text{9}}\text{g coal}\times \frac{\text{0}\text{.9}\text{g}\text{S}}{\text{100}\text{g}\text{coal}}\\ =2.61\times {10}^7\text{g}\text{S}\end{array}$

Reaction between sulfur and oxygen to produce sulfur dioxide can be given as shown.

    $\text{S}+{\text{O}}_{\text{2}}\to {\text{SO}}_{\text{2}}$

From the equation, it is found that one mole of sulfur reacts with one mole of oxygen to form one mole of sulfur dioxide.  Molar mass of sulfur is $\text{32}\text{.1}\text{g}$ and that of ${\text{SO}}_{\text{2}}$ is $\text{64}\text{.1}\text{g}$.  This means $\text{32}\text{.1}\text{g}$ of sulfur burns in oxygen to produce $\text{64}\text{.1}\text{g}$ of sulfur dioxide.  Therefore, the amount of sulfur dioxide produced by burning $2.61\times {10}^7\text{g}$ of sulfur can be calculated as shown below.

    $\begin{array}{c}2.61\times {10}^7\text{g}\text{S×}\frac{\text{1}\text{mol}\text{S}}{\text{32}\text{.1}\text{g}\text{S}}\text{×}\frac{\text{1}\text{mol}{\text{SO}}_{\text{2}}}{\text{1}\text{mol}\text{S}}\text{×}\frac{\text{64}\text{.1}\text{g}{\text{SO}}_{\text{2}}}{\text{1}\text{mol}{\text{SO}}_{\text{2}}}=\frac{\text{167}\text{.301}\times {10}^7}{32.1}\text{g}{\text{SO}}_{\text{2}}\\ =\text{5}\text{.2}\times {10}^7\text{g}{\text{SO}}_{\text{2}}\end{array}$

Conversion of gram into metric ton can be done using the conversion factor as shown below.

    $\begin{array}{c}\text{5}\text{.2}\times {10}^7\text{g}{\text{SO}}_{\text{2}}=\text{5}\text{.2}\times {10}^7\text{g}\times \frac{\text{1}\text{metric}\text{ton}}{\text{1000000}\text{g}}\\ =\text{5}\text{.2}\times {10}^1\text{metric}\text{ton}\\ =\text{52 metric}\text{ton}\end{array}$

Therefore, the amount of sulfur dioxide produced is $\text{52 metric}\text{tons}$.