Chapter 5, Problem 5.110P

(a)

Interpretation Introduction

Interpretation:

The grams of nickel that can be converted to the carbonyl with $3.55{\text{ m}}^{\text{3}}$ at $100.7\text{ kPa}$ is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

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 5.110P

The grams of nickel that can be converted to the carbonyl with $3.55{\text{ m}}^{\text{3}}$ at $100.7\text{ kPa}$ is $1.95\times {10}^3\text{ g}$.

Explanation of Solution

The equation for the reaction of $\text{Ni}\left(aq\right)$ with $\text{CO}\left(g\right)$ is as follows:

$\text{Ni}\left(s\right)\text{ + 4CO}\left(aq\right)\to \text{Ni}{\left(\text{CO}\right)}_4\left(g\right)$        (1)

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

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

Substitute $\text{50 }°\text{C}$ for $T\left(\text{°C}\right)$ in equation (2).

$\begin{array}{c}T\left(\text{K}\right)=50+273.15\\ =323.15\text{ K}\end{array}$

The expression to convert pressure from $\text{kPa}$ into $\text{atm}$ is as follows,

$\begin{array}{c}P=\text{ 700 torr}\left(\frac{1\text{ atm}}{100.7\text{ kPa}}\right)\\ =0.993831729\text{ atm}\end{array}$

The expression to calculate the moles of the $\text{CO}\left(g\right)$ is as follows,

$PV=nRT$        (3)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of $\text{CO}\left(g\right)$ and $R$ is the gas constant.

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

$n=\frac{PV}{RT}$        (4)

Substitute the value $0.993831729\text{ atm}$ for $P$, $323.15\text{ K}$ for $T$, $\text{3}{\text{.55 m}}^{\text{3}}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (4).

$\begin{array}{c}n=\frac{\left(0.993831729\text{ atm}\right)\left(\text{3}{\text{.55 m}}^{\text{3}}\right)\left(\frac{1000\text{ L}}{{\text{1 m}}^{\text{3}}}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{323}\text{.15 K}\right)}\\ =1951.183\text{ mol}\end{array}$

From equation (1), four moles of the $\text{CO}\left(g\right)$ reacts with one mole of $\text{Ni}\left(s\right)$. Thus, the moles of $\text{Ni}\left(s\right)$ is calculated from $\text{CO}\left(g\right)$ as follows:

$\text{Moles of Ni}\left(s\right)=\left[\left(\text{Moles of CO}\left(g\right)\right)\left(\frac{\text{1 mol Ni}\left(s\right)}{\text{4 mol CO}\left(g\right)}\right)\right]$        (5)

Substitute the value $1951.183\text{ mol}$ for moles of $\text{CO}\left(g\right)$ in the equation (5).

$\begin{array}{c}\text{Moles of Ni}\left(s\right)=\left[\left(1951.183\text{ mol}\right)\left(\frac{\text{1 mol Ni}\left(s\right)}{\text{4 mol CO}\left(g\right)}\right)\right]\\ =487.79\text{ mol}\end{array}$

The expression to calculate the mass of $\text{Ni}\left(s\right)$ is as follows:

$\text{Moles of Ni}=\frac{\text{Mass of Ni}}{\text{Molar mass of Ni}}$        (6)

Rearrange the equation (6) to calculate the mass of $\text{Ni}\left(s\right)$ is as follows:

$\text{Mass of Ni}=\left(\text{Moles of Ni}\right)\left(\text{Molar mass of Ni}\right)$        (7)

Substitute the value $487.79\text{ mol}$ for moles of $\text{Ni}\left(s\right)$ and $58.69\text{ g/mol}$ for molar mass of $\text{Ni}\left(s\right)$ in the equation (7).

$\begin{array}{c}\text{Mass of Ni}=\left(487.79\text{ mol}\right)\left(58.69\text{ g/mol}\right)\\ =1951.183\text{ g}\\ \approx 1.95\times {10}^3\text{ g}\end{array}$

Conclusion

The grams of nickel that can be converted to the carbonyl with $3.55{\text{ m}}^{\text{3}}$ at $100.7\text{ kPa}$ is $1.95\times {10}^3\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

The mass of nickel obtained per cubic meter of the carbonyl at $\text{155 }°\text{C}$ and $21\text{ atm}$ is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

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 5.110P

The mass of nickel obtained per cubic meter of the carbonyl at $\text{155 }°\text{C}$ and $21\text{ atm}$ is $3.5\times {10}^4\text{ g}$.

Explanation of Solution

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

$T\left(\text{K}\right)=T\left(\text{°C}\right)+273.15$        (8)

Substitute $\text{155 }°\text{C}$ for $T\left(\text{°C}\right)$ in equation (8).

$\begin{array}{c}T\left(\text{K}\right)=155°\text{C}+273.15\\ =428.15\text{ K}\end{array}$

The expression to calculate the moles of the $\text{Ni}{\left(\text{CO}\right)}_4$ is as follows,

$PV=nRT$        (9)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of $\text{Ni}{\left(\text{CO}\right)}_4$ and $R$ is the gas constant.

Rearrange the equation (9) to calculate $n$ as follows,

$n=\frac{PV}{RT}$        (10)

Substitute the value $\text{21 atm}$ for $P$, $\text{428}\text{.15 K}$ for $T$, ${\text{1 m}}^{\text{3}}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (10).

$\begin{array}{c}n=\frac{\left(\text{21 atm}\right)\left({\text{1 m}}^{\text{3}}\right)\left(\frac{1000\text{ L}}{{\text{1 m}}^{\text{3}}}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{428}\text{.15 K}\right)}\\ =597.42059\text{ mol}\end{array}$

From the equation (1), one mole of the $\text{Ni}\left(s\right)$ will be formed one mole of $\text{Ni}{\left(\text{CO}\right)}_4$. Thus, the moles of $\text{Ni}\left(s\right)$ is calculated from $\text{Ni}{\left(\text{CO}\right)}_4$ as follows:

$\text{Moles of Ni}\left(s\right)=\left[\left(\text{Moles of Ni}{\left(\text{CO}\right)}_4\right)\left(\frac{\text{1 mol Ni}\left(s\right)}{\text{2 mol Ni}{\left(\text{CO}\right)}_4}\right)\right]$        (11)

Substitute the value $597.42059\text{ mol}$ for moles of $\text{CO}\left(g\right)$ in the equation (11).

$\begin{array}{c}\text{Moles of Ni}\left(s\right)=\left[\left(597.42059\text{ mol}\right)\left(\frac{\text{1 mol Ni}\left(s\right)}{\text{1 mol Ni}{\left(\text{CO}\right)}_4\left(g\right)}\right)\right]\\ =597.42059\text{ mol}\end{array}$

The expression to calculate the mass of $\text{Ni}\left(s\right)$ is as follows:

$\text{Moles of Ni}=\frac{\text{Mass of Ni}}{\text{Molar mass of Ni}}$        (12)

Rearrange the equation (12) to calculate the mass of $\text{Ni}\left(s\right)$ is as follows:

$\text{Mass of Ni}=\left(\text{Moles of Ni}\right)\left(\text{Molar mass of Ni}\right)$        (13)

Substitute the value $597.42059\text{ mol}$ for moles of $\text{Ni}\left(s\right)$ and $58.69\text{ g/mol}$ in the equation (13).

$\begin{array}{c}\text{Mass of Ni}=\left(597.42059\text{ mol}\right)\left(58.69\text{ g/mol}\right)\\ =3.50626\times {10}^4\text{ g}\\ \approx 3.5\times {10}^4\text{ g}\end{array}$

Conclusion

The mass of nickel obtained per cubic meter of the carbonyl at $\text{155 }°\text{C}$ and $21\text{ atm}$ is $3.5\times {10}^4\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

The volume of $\text{CO}\left(g\right)$ at $35°\text{C}$ and $769\text{ torr}$ is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

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 5.110P

The volume of $\text{CO}\left(g\right)$ formed at $35°\text{C}$ and $769\text{ torr}$ is $63{\text{ m}}^3$.

Explanation of Solution

The expression to calculate the mass of $\text{Ni}\left(s\right)$ is as follows:

$\text{Moles of Ni}=\frac{\text{Mass of Ni}}{\text{Molar mass of Ni}}$        (14)

Substitute $3.50626\times {10}^4\text{ g}$ for the mass of $\text{Ni}\left(s\right)$ and $58.69\text{ g/mol}$ in the equation (14).

$\begin{array}{c}\text{Moles of Ni}=\frac{\left(3.50626\times {10}^4\text{ g}\right)}{\left(58.69\text{ g/mol}\right)}\\ =597.420\text{ mol}\end{array}$

From the equation (1), four moles of the $\text{CO}\left(g\right)$ reacts with one mole of $\text{Ni}\left(s\right)$. Thus, the moles of $\text{CO}\left(g\right)$ is calculated from $\text{Ni}\left(s\right)$ as follows:

$\text{Moles of CO}\left(g\right)=\left[\left(\text{Moles of CO}\left(g\right)\right)\left(\frac{\text{1 mol CO}\left(g\right)}{\text{4 mol Ni}\left(s\right)}\right)\right]$        (15)

Substitute the value $597.420\text{ mol}$ for moles of $\text{CO}\left(g\right)$ in the equation (15).

$\begin{array}{c}\text{Moles of CO}\left(g\right)=\left[\left(597.420\text{ mol}\right)\left(\frac{\text{4 mol CO}\left(g\right)}{\text{1 mol Ni}\left(s\right)}\right)\right]\\ =2389.68238\text{ mol}\end{array}$

Substitute $\text{35 }°\text{C}$ for $T_{\text{°C}}$ in equation (8).

$\begin{array}{c}T\left(\text{K}\right)=\text{35 }°\text{C}+273.15\\ =308.15\text{ K}\end{array}$

The expression to calculate the pressure of $\text{CO}\left(g\right)$ is as follows,

$P_{\text{total}}=P_{\text{water vapour}}+P_{\text{CO}}$        (16)

Rearrange the equation (16) to calculate the $P_{\text{CO}}$.

$P_{\text{CO}}=P_{\text{total}}-P_{\text{water vapour}}$        (17)

Substitute $42.2\text{ torr}$ for $P_{\text{water vapour}}$ and $769\text{ torr}$ for $P_{\text{total}}$ in the equation (17).

$\begin{array}{c}{P}_{\text{CO}}=769\text{ torr}-42.2\text{ torr}\\ =\text{726}\text{.8 torr}\end{array}$

The expression to convert $\text{torr}$ into $\text{atm}$ is as follows,

$\begin{array}{c}P=726.8\text{ torr}\left(\frac{1\text{ atm}}{\text{760 torr}}\right)\\ =0.956315789\text{ atm}\end{array}$

The expression to calculate the volume of $\text{CO}\left(g\right)$ is as follows,

$PV=nRT$        (18)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of $\text{CO}\left(g\right)$ and $R$ is the gas constant.

Rearrange the equation (18) to calculate $V$ as follows,

$V=\frac{nRT}{P}$        (19)

Substitute the value $0.993831729\text{ atm}$ for $P$, $308.15\text{ K}$ for $T$, $2389.68238\text{ mol}$ for $n$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (4).

$\begin{array}{c}V=\frac{\left(2389.68238\text{ mol}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(308.15\text{ K}\right)}{\left(0.956315789\text{ atm}\right)}\\ =\left(63218.4995\text{ L}\right)\left(\frac{1{\text{ m}}^{\text{3}}}{1000\text{ L}}\right)\\ =63.2184995{\text{ m}}^{\text{3}}\\ \approx 63{\text{ m}}^{\text{3}}\end{array}$

Conclusion

The volume of $\text{CO}\left(g\right)$ at $\text{35 }°\text{C}$ and $\text{769 torr}$ is $63{\text{ m}}^{\text{3}}$.