Chapter 9, Problem 136QP

Interpretation Introduction

Interpretation:

The volume of ${\text{CO}}_2$ gas released into the atmosphere daily by the company is to be determined.

Concept Introduction:

The number of moles of a gas $\left(n\right)$ is equal to the ratio of amount of gas $\left(m\right)$ present in the container to the molar mass $\left(M\right)$ of that gas.

$n=\frac{m}{M}$ … (1)

Ideal gas law is the law followed by ideal gases. It states that pressure $\left(P\right)$ and volume $\left(V\right)$ of the gas is directly proportional to the number of moles $\left(n\right)$ present in the gas and the temperature $\left(T\right)$ of the gas.

$\begin{array}{l}PV\propto nT\\ PV=nRT\end{array}$ … (2)

Where, $R$ is universal gas constant.

Answer to Problem 136QP

Solution:

The volume of ${\text{CO}}_2$ gas released into the atmosphere daily by the company is $2.53\times {10}^7\text{L}$ .

Explanation of Solution

Given Information: The amount of limestone used daily is $1.00\times {10}^5\text{ kg}$ at $735\text{ torr}$ and $25.0°\text{C}$ .

The reaction is as follows,

${\text{CaCO}}_3\left(s\right)\stackrel{\text{heat}}{\to }\text{CaO}\left(s\right)+{\text{CO}}_2\left(g\right)$

The number of moles of ${\text{CaCO}}_3$ decomposed upon heating is determined by using equation (1). The molar mass of ${\text{CaCO}}_3$ is,

$\begin{array}{c}{M}_{{\text{CaCO}}_3}=40.09{\text{ g mol}}^{-1}+12{\text{ g mol}}^{-1}+16\times 3{\text{ g mol}}^{-1}\\ =100.09{\text{ g mol}}^{-1}\end{array}$

Now, substitute $M$ as $\text{100}{\text{.09 g mol}}^{-1}$ and $m$ as $\text{1}\text{.00}\times {\text{10}}^8\text{ g}$ in equation (1) to determine $n$ .

$\begin{array}{c}n=\frac{\text{1}\text{.0}\times {\text{10}}^8\text{ g}}{\text{100}{\text{.09 g mol}}^{-1}}\\ =9.99\times {10}^5\text{ mol}\end{array}$

According to the reaction,

$\begin{array}{c}1{\text{ mol CaCO}}_3\text{ produces}=1{\text{ mol CO}}_2\\ 9.99\times {10}^5{\text{ mol CaCO}}_3\text{ produces}=\frac{1{\text{ mol CO}}_2}{1{\text{ mol CaCO}}_3}\times 9.99\times {10}^5{\text{ mol CaCO}}_3\\ =9.99\times {10}^5{\text{ mol CO}}_2\end{array}$

Therefore, $9.99\times {10}^5{\text{ mol CO}}_2$ is produced by $1.00\times {10}^5\text{ kg}$ of limestone at $735\text{ torr}$ and $25.0°\text{C}$ . So, the volume of ${\text{CO}}_2$ is determined by equation (2).

Convert degree Celsius temperature into Kelvin.

$T\text{ K}=T°\text{C}+273.15$

Temperature $25°\text{C}$ is converted into Kelvin.

$\begin{array}{c}T=25.0°\text{C}+273.15\\ =298.2\text{ K}\end{array}$

Convert pressure units from $\text{torr}$ to $\text{atm}$ .

$\begin{array}{l}1\text{ atm}=760\text{ torr}\\ \text{1 torr}=\frac{1}{760}\text{ atm}\end{array}$

For $735\text{ torr}$ , pressure $P$ is,

$\begin{array}{c}\text{1 torr}=\frac{1}{760}\text{ atm}\\ 735\text{ torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 735\text{ torr}\\ =0.967\text{ atm}\end{array}$

Substitute $P$ as $0.967\text{ atm}$ , $T$ as $298.2\text{ K}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ , and $n$ as $9.99\times {10}^5\text{ mol}$ in equation (2) to determine volume $\left(V\right)$ of ${\text{H}}_2$ gas.

$\begin{array}{c}0.967\text{ atm}\times V=9.99\times {10}^5\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 298.2\text{ K}\\ V=\frac{9.99\times {10}^5\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 298.2\text{ K}}{0.967\text{ atm}}\\ =2.53\times {10}^7\text{L}\end{array}$

Conclusion

The volume of ${\text{CO}}_2$ gas released into the atmosphere daily by the company is $2.53\times {10}^7\text{L}$ .