Chapter 5, Problem 86P

5-114 Carbon dioxide gas, saturated with water vapor, can be produced by the addition of aqueous acid to calcium carbonate based on the following balanced net ionic equation:

(a) How many moles of wet CO (g), collected at 60.°C and 774 torr total pressure, are produced by the complete reaction of 10.0 g of CaCO₃ with excess acid?

(b) What volume does this wet CO₂ occupy?

(c) What volume would the CO₂ occupy at 774 torr if a desiccant (a chemical drying agent) were added to remove the water? The vapor pressure of water at 60.°C is 149.4 mm Hg.

Interpretation Introduction

(a)

Interpretation:

The number of moles of wet ${\text{CO}}_{\text{2}}\left(g\right)$ collected at ${\text{60 }}^{\text{o}}\text{C}$ and $774\text{ torr}$ total pressure.

Concept Introduction:

Moles of wet ${\text{CO}}_{\text{2}}\left(g\right)$ collected can be calculated from the balanced chemical equation.

${\text{1 mole of CaCO}}_{\text{3}}{\text{= 1 mole of CO}}_{\text{2}}$.

Answer to Problem 86P

The number of moles of carbon dioxide produced is $\text{0}\text{.10 mol}$.

Explanation of Solution

The weight of calcium carbonate is $10\text{ g}$

The temperature is $\text{60 }°\text{C}$

The pressure is $774\text{ torr}$

The molecular weight of calcium carbonate is calculated below:

$\begin{array}{c}M=40+12+3\times 16\\ =100\text{ g/mol}\end{array}$

The number of moles in $10\text{ g}$ of calcium carbonate is calculated below.

$\begin{array}{c}=\frac{10\text{g}}{100\text{ g/mol}}\\ =0.10\text{mol}\end{array}$

The balanced chemical equation is below:

${\text{CaCO}}_{\text{3}}\left(s\right)+2{\text{H}}^{+}\left(\text{a}q\right) \to {\text{Ca}}^{2+}\left(\text{a}q\right) + {\text{H}}_{\text{2}}\text{O}(\text{l}) + {\text{CO}}_{\text{2}}\left(g\right)$

From the above equation,

${\text{1 mole of CaCO}}_{\text{3}}\text{= 1 mole of}\text{C}{\text{O}}_2$

Therefore, the number of moles of carbon dioxide produced is $0.10\text{ mol}$.

Interpretation Introduction

(b)

Interpretation:

The volume of wet ${\text{CO}}_{\text{2}}$ occupy should be calculated.

Concept Introduction:

To determine the volume of wet ${\text{CO}}_{\text{2}}$ occupy can be determined from following relationship.

The relationship between pressure, volume, and temperature, under two sets of conditions, is given by following equation.

$\frac{P_1{V}_1}{T_1}=\frac{P_0{V}_0}{T_0}$.

Answer to Problem 86P

The value of carbon dioxide produced is $2.68\text{ L}$.

Explanation of Solution

The weight of calcium carbonate is $10\text{ g}$

The temperature is.

$60\text{ °C}$

The pressure is $774\text{ torr}$

The molecular weight of calcium carbonate is calculated below:

$\begin{array}{l}=40+12+3\times 16\hfill \\ =100\text{ g/mol}\hfill \end{array}$

The number of moles in $10\text{ g}$ of calcium carbonate is calculated below:

$\begin{array}{l}=\frac{10\text{g}}{100\text{g/mol}}\\ =0.10\text{mol}\end{array}$

The balanced chemical equation is below:

${\text{CaCO}}_{\text{3}}\left(s\right)+ 2{\text{H}}^{+}\left(\text{a}q\right) \to {\text{ Ca}}^{2+}\left(\text{a}q\right) +{\text{ H}}_{\text{2}}\text{O}(\text{l}) +{\text{ CO}}_2\left(g\right)$

From the above equation:

${\text{1 mole of CaCO}}_{\text{3}}{\text{= 1 mole of CO}}_{\text{2}}$

Therefore, the number of moles of carbon dioxide produced is $0.10\text{ mol}$

We know that at STP, 1 mole of any gas occupies $22.4\text{ L}$

Therefore,

$\begin{array}{c}{\text{1 mole of CO}}_{\text{2}}=22.4\text{ L at STP}\\ \text{0}{\text{.1 mole of CaCO}}_3=\text{0}{\text{.1 mole of CO}}_{\text{2}}\\ =\text{2}{\text{.24 L of CO}}_2\text{at STP}\end{array}$

Therefore, the volume of $0.10\text{ mol}$ of carbon dioxide at STP is calculated below:

$\begin{array}{l}=\text{0}\text{.10 mol}\text{×}\frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mol}}\\ =\text{2}\text{.24 L}\end{array}$

At STP, the value of $P_0$, $T_0$, $V_0$

$\begin{array}{l}{P}_0=760\text{ mm Hg}\\ T_0=273\text{ K}\\ V_0=2.24\text{ L}\end{array}$

Now, convert the pressure from $\text{torr}$ to $\text{mm Hg}$

We know that,

$\text{1 torr }=\text{1mm Hg}$

Therefore, the pressure is calculated below:

$\begin{array}{c}{\text{P}}_{\text{1}}=\text{774 torr×}\frac{\text{1 mm Hg}}{\text{1 torr}}\\ =\text{774 mm Hg}\end{array}$

Now, convert temperature from $\text{°C}$ to K.

We know that,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$

Therefore,

$\begin{array}{c}{T}_1=60\text{ °C}+273\\ =333\text{ K}\end{array}$

The relationship between pressure, volume, and temperature, under two sets of conditions, is given by following equation:

$\frac{P_1{V}_1}{T_1}=\frac{P_0{V}_0}{T_0}$

Upon rearranging, we get,

$V_1=\frac{P_0{V}_0}{T_0}\times \frac{T_1}{P_1}$

By substituting the values of $P_0$, $V_0$, $T_0$ as well as $P_1$ and $T_1$ in the above equation, we can calculate the value of $V_1$

$\begin{array}{l}{V}_1=\frac{\text{760}\text{mm}\text{Hg}\text{×}\text{2}\text{.24}\text{L}}{\text{273}\text{K}}\text{×}\frac{\text{333}\text{K}}{\text{774}\text{mm}\text{Hg}}\\ =\text{2}\text{.68}\text{L}\end{array}$

Hence the value of carbon dioxide produced is $\text{2}\text{.68 L}$.

Interpretation Introduction

(c)

Interpretation:

The volume of wet ${\text{CO}}_{\text{2}}$ occupy at $774\text{ torr}$

Concept Introduction:

To determine the volume of wet ${\text{CO}}_{\text{2}}$ occupy can be determined from following relationship.

The relationship between pressure, volume, and temperature, under two sets of conditions, is given by following equation.

$\frac{P_1{V}_1}{T_1}=\frac{P_0{V}_0}{T_0}$.

Answer to Problem 86P

$\text{3}\text{.32 L}$.

Explanation of Solution

The weight of calcium carbonate is $10\text{ g}$

The temperature is $60\text{ °C}$

The pressure is $774\text{ torr}$

The vapor pressure of water at $60\text{ °C}$ is $149.4\text{ mm Hg}$

The molecular weight of calcium carbonate is calculated below:

$\begin{array}{l}=40+ 12+3\times 16\hfill \\ =100\text{ g/mol}\hfill \end{array}$

The number of moles in $10\text{ g}$ of calcium carbonate is calculated below:

$\begin{array}{l}=\frac{10\text{g}}{100\frac{\text{g}}{\text{mol}}}\\ =0.10\text{mol}\end{array}$

The balanced chemical equation is below:

${\text{CaCO}}_{\text{3}}\left(s\right)+ 2{\text{H}}^{+}\left(\text{a}q\right) \to {\text{ Ca}}^{2+}\left(\text{a}q\right) +{\text{ H}}_{\text{2}}\text{O}(\text{l}) +{\text{ CO}}_2\left(g\right)$

From the above equation:

${\text{1 mole of CaCO}}_{\text{3}}{\text{= 1 mole of CO}}_{\text{2}}$

Therefore, the number of moles of carbon dioxide produced is $0.10\text{ mol}$

We know that at STP, 1 mole of any gas occupies $22.4\text{ L}$

Therefore,

$\begin{array}{c}{\text{1 mole of CO}}_{\text{2}}=22.4\text{ L at STP}\\ \text{0}{\text{.1 mole of CaCO}}_3=\text{0}{\text{.1 mole of CO}}_{\text{2}}\\ =\text{2}{\text{.24 L of CO}}_2\text{at STP}\end{array}$

Therefore, the volume of $0.10\text{ mol}$ of carbon dioxide at STP is calculated below:

$\begin{array}{l}=\text{0}\text{.10 mol}\text{×}\frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mol}}\\ =\text{2}\text{.24 L}\end{array}$

At STP, the value of $P_0$, $T_0$, $V_0$

$\begin{array}{l}{P}_0=760\text{ mm Hg}\\ T_0=273\text{ K}\\ V_0=2.24\text{ L}\end{array}$

The pressure is $774\text{ torr}$

Now, convert the pressure from $\text{torr}$ to $\text{mm Hg}$ ,

We know that,

$\text{1 torr = 1 mm Hg}$

Therefore, the pressure is calculated below,

$\begin{array}{c}{P}_1=774\text{ torr}\times \frac{1\text{ mm Hg}}{1\text{ torr}}\\ =774\text{ mm Hg}\end{array}$

The vapor pressure of water at $60\text{ °C}$ is $149.4\text{ mm Hg}$. When desiccator is used, it will absorb the water vapor.

Therefore, the pressure is calculated below,

$\begin{array}{c}{P}_1=774\text{ mm Hg }-149.4\text{ mm Hg}\\ =624.6\text{ mm Hg}\end{array}$

Now, convert temperature from°C to K.

We know that,

$T\left(K\right)=T\left(°\text{C}\right)+273$

Therefore,

$\begin{array}{c}{\text{T}}_{\text{1}}\text{=60° C + 273}\\ \text{=333 K}\end{array}$

The relationship between pressure, volume, and temperature, under two sets of conditions, is given by following equation,

$\frac{P_1{V}_1}{T_1}=\frac{P_0{V}_0}{T_0}$

Upon rearranging, we get,

$V_1=\frac{P_0{V}_0}{T_0}\times \frac{T_1}{P_1}$

By substituting the values of $P_0$, $V_0$, $T_0$ as well as $P_1$ and $T_1$ in the above equation, we can calculate the value of $V_1$

$\begin{array}{c}{V}_1=\frac{760\text{mm}\text{Hg}\times 2.24\text{L}}{273\text{K}}\times \frac{333\text{K}}{624.6\text{mm}\text{Hg}}\\ =3.32\text{L}\end{array}$

Hence the value of carbon dioxide produced is $3.32\text{ L}$.