Chapter 5, Problem 5.139P

(a)

Interpretation Introduction

Interpretation:

The mass of calcium sulfate from the scrubbing $\text{4 GL}$ of flue gas is to be calculated.

Concept introduction:

The expression to calculate the partial pressure of the gas is as follows:

$P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ is the mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows,

$X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.139P

The mass of calcium sulfate from the scrubbing $\text{4 GL}$ of flue gas is $\text{4 kg}$.

Explanation of Solution

The concentration of ${\text{SO}}_{\text{2}}$ is $1000$ times higher than its mole fraction in clear dry air that is $2\times {10}^{-10}$. Thus, the mole fraction of ${\text{SO}}_{\text{2}}$ equals to $2\times {10}^{-7}$.

The expression to calculate the volume of ${\text{SO}}_{\text{2}}$ is as follows:

${\text{Volume of SO}}_{\text{2}}\text{ removed}=\left[\begin{array}{l}\left({\text{mole fraction of SO}}_{\text{2}}\right)\left(\text{volume of flue gas}\right)\\ \left({\text{percentage of SO}}_{\text{2}}\text{ removes}\right)\end{array}\right]$        (1)

Substitute the value $2\times {10}^{-7}$ for mole fraction of ${\text{SO}}_{\text{2}}$, $4\text{ GL}$ for the volume of flue gas and $95\text{ \%}$ for a percentage of ${\text{SO}}_{\text{2}}$ removes in the equation (1).

$\begin{array}{c}{\text{Volume of SO}}_{\text{2}}\text{ removed}=\left[\begin{array}{l}\left(2\times {10}^{-7}\right)\left(4\text{ GL}\left(\frac{{10}^9\text{ L}}{1\text{ GL}}\right)\right)\\ \left(\frac{95\text{ \%}}{100\text{ \%}}\right)\end{array}\right]\\ =760\text{ L}\end{array}$

The formula to convert $°\text{C}$ to Kelvin is:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$        (2)

Substitute $\text{25 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (5).

$\begin{array}{c}T\left(\text{K}\right)=\text{25 }°\text{C}+273.15\\ =298.15\text{ K}\end{array}$

The expression to calculate the moles of the ${\text{SO}}_{\text{2}}$ is as follows,

$PV=nRT$        (3)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of ${\text{SO}}_{\text{2}}$ and $R$ is the gas constant.

Rearrange the equation (3) to calculate $n$ as follows,

$n=\frac{PV}{RT}$        (4)

Substitute the value $\text{1}\text{.0 atm}$ for $P$, $298.15\text{ K}$ for $T$, $\text{760 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (4).

$\begin{array}{c}n=\frac{\left(\text{760 L}\right)\left(\text{1}\text{.0 atm}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(298.15\text{ K}\right)}\\ =31.04814\text{ mol}\end{array}$

The equation for the reaction of ${\text{CaCO}}_{\text{3}}$ with ${\text{SO}}_{\text{2}}$ is as follows:

${\text{CaCO}}_{\text{3}}\left(s\right){\text{ + SO}}_{\text{2}}\left(g\right)\to {\text{CaSO}}_{\text{3}}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$        (5)

The equation for the reaction of ${\text{CaSO}}_{\text{3}}$ with ${\text{O}}_{\text{2}}$ is as follows:

${\text{2CaSO}}_{\text{3}}\left(s\right){\text{ + O}}_{\text{2}}\left(g\right)\to 2{\text{CaSO}}_4\left(s\right)$        (6)

From the equation (5), one mole of the ${\text{SO}}_{\text{2}}$ will be formed one mole of ${\text{CaSO}}_{\text{3}}$. From the equation (6), two moles of the ${\text{CaSO}}_{\text{3}}$ will be formed two moles of ${\text{CaSO}}_4$. Thus, the moles of ${\text{CaSO}}_4$ is calculated from ${\text{SO}}_{\text{2}}$ as follows:

${\text{Moles of CaSO}}_4=\left[\left({\text{Moles of SO}}_2\right)\left(\frac{{\text{1 mol CaSO}}_4}{{\text{1 mol SO}}_2}\right)\right]$        (7)

Substitute the value $31.04814\text{ mol}$ for moles of ${\text{SO}}_{\text{2}}$ in the equation (7).

$\begin{array}{c}{\text{Moles of CaSO}}_4=\left[\left(31.04814\text{ mol}\right)\left(\frac{{\text{1 mol CaSO}}_4}{{\text{1 mol SO}}_2}\right)\right]\\ =31.04814\text{ mol}\end{array}$

The expression to calculate the mass of ${\text{CaSO}}_4$ is as follows:

${\text{Moles of CaSO}}_{\text{4}}=\frac{{\text{Mass of CaSO}}_{\text{4}}}{{\text{Molar mass CaSO}}_{\text{4}}}$        (8)

Rearrange the expression to calculate the mass of ${\text{CaSO}}_4$ in the equation (8)

${\text{Mass of CaSO}}_{\text{4}}=\left({\text{Moles of CaSO}}_{\text{4}}\right)\left({\text{Molar mass CaSO}}_{\text{4}}\right)$        (9)

Substitute the value $31.04814\text{ mol}$ for the mass of ${\text{CaSO}}_4$ and $136.14\text{ g/mol}$ for the molar mass of ${\text{CaSO}}_4$ in the equation (9).

$\begin{array}{c}{\text{Mass of CaSO}}_{\text{4}}=\left(31.04814\text{ mol}\right)\left(136.14\text{ g/mol}\right)\left(\frac{1\text{ kg}}{{10}^3\text{g}}\right)\\ =4.2269\text{ kg}\\ =\text{4 kg}\end{array}$

Conclusion

The mass of calcium sulfate from the scrubbing $\text{4 GL}$ of flue gas is $\text{4 kg}$.

(b)

Interpretation Introduction

Interpretation:

The volume of air at $\text{25 }°\text{C}$ and $1.0\text{ atm}$ is needed to react with all calcium sulfite is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.139P

The volume of air at $\text{25 }°\text{C}$ and $1.0\text{ atm}$ is $\text{2000 L}$.

Explanation of Solution

From the equation (6), one mole of the ${\text{O}}_2$ will be formed two moles of ${\text{CaSO}}_4$. Thus, the moles of ${\text{O}}_2$ is calculated from ${\text{CaSO}}_4$ as follows:

${\text{Moles of O}}_2=\left[\left({\text{Moles of CaSO}}_4\right)\left(\frac{{\text{1 mol O}}_2}{{\text{2 mol CaSO}}_4}\right)\right]$        (10)

Substitute the value $31.04814\text{ mol}$ for moles of ${\text{CaSO}}_4$ in the equation (10).

$\begin{array}{c}{\text{Moles of O}}_2=\left[\left(31.04814\text{ mol}\right)\left(\frac{{\text{1 mol O}}_2}{{\text{2 mol CaSO}}_4}\right)\right]\\ =15.52407\text{ mol}\end{array}$

The formula to convert $°\text{C}$ to Kelvin is:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$        (11)

Substitute $\text{25 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (11).

$\begin{array}{c}T\left(\text{K}\right)=\text{25 }°\text{C}+273.15\\ =298.15\text{ K}\end{array}$

The expression to calculate the volume of ${\text{O}}_2$ is as follows,

$PV=nRT$        (12)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the ${\text{O}}_2$ and $R$ is the gas constant.

Rearrange the equation (7) to calculate $V$ as follows,

$V=\frac{nRT}{P}$        (13)

Substitute the value $\text{1}\text{.0 atm}$ for $P$, $298.15\text{ K}$ for $T$, $15.52407\text{ mol}$ for $n$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (13).

$\begin{array}{c}V=\frac{\left(15.52407\text{ mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(298.15\text{ K}\right)}{1\text{ atm}}\\ =\text{380 L}\end{array}$

The expression to calculate the volume of air is as follows:

$\text{Volume of air}=\frac{\left({\text{Volume of O}}_{\text{2}}\right)}{\left({\text{Mole fraction of O}}_{\text{2}}\right)}$        (14)

Substitute the value $0.209$ for mole fraction of ${\text{O}}_2$, $\text{380 L}$ for the volume of ${\text{O}}_2$ in the equation (14).

$\begin{array}{c}\text{Volume of air}=\frac{\left(380\text{ L}\right)}{\left(0.209\right)}\\ =1818.18\text{ L}\\ \approx \text{2000 L}\end{array}$

Conclusion

The volume of air at $\text{25 }°\text{C}$ and $1.0\text{ atm}$ is $\text{2000 L}$.