Chapter 5, Problem 152CP

Interpretation Introduction

Interpretation:Mass percents of $\text{CaO}$ and $\text{BaO}$ in mixture are to be determined.

Concept introduction:State of any gas is determined by some of its parameters. These parameters are pressure, amount of gas, temperature and volume. Ideal gas law helps to govern state of gas with help of relationships between these gas parameters.

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Answer to Problem 152CP

Mass percent of $\text{BaO}$ is $13.38\text{ \%}$ and that of $\text{CaO}$ is $86.61\text{ \%}$ .

Explanation of Solution

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation for $n$ .

  $n=\frac{PV}{RT}$

Temperature is calculated as follows:

  $\begin{array}{c}T=\left(30.0+273\right)\text{ K}\\ =\text{303 K}\end{array}$

Value of $V$ is $\text{1}\text{.50 L}$ .

Value of $R$ is $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $T$ is $303\text{ K}$ .

Value of $P$ is $750\text{ torr}$ .

Substitute the values in above equation for initial moles of ${\text{CO}}_2$ .

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(750\text{ torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(1.50\text{ L}\right)}{\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(303\text{ K}\right)}\\ =0.0595\text{ mol}\end{array}$

Value of $V$ is $\text{1}\text{.50 L}$ .

Value of $R$ is $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $T$ is $303\text{ K}$ .

Value of $P$ is $230\text{ torr}$ .

Substitute the values in above equation for final moles of ${\text{CO}}_2$ .

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(230\text{ torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\left(1.50\text{ L}\right)}{\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(303\text{ K}\right)}\\ =0.0183\text{ mol}\end{array}$

Number of moles of ${\text{CO}}_2$ consumed is calculated as follows:

  $\begin{array}{c}{\text{Number of moles of CO}}_2\text{ consumed}=\left(0.0595-0.0183\right)\text{ mol}\\ =\text{0}\text{.0412 mol}\end{array}$

Consider $x\text{ g}$ to be mass of $\text{BaO}$ . Total mass of mixture of $\text{BaO}$ and $\text{CaO}$ is $5.14\text{ g}$ . So mass of $\text{CaO}$ is calculated as follows:

  $\text{Mass of CaO}\left(=5.14-x\right)\text{ g}$

Expression for moles of substance is as follows:

  $n=\frac{m}{M}$

Where,

• $n$ is amount or moles of substance.
• $m$ is mass of substance.
• $M$ is molar mass of substance.

Value of $m$ is $x\text{ g}$ .

Value of $M$ is $56.08\text{ g/mol}$ .

Substitute the values in above equation for moles of $\text{BaO}$ .

  $\text{Moles of BaO}=\frac{x\text{ g}}{56.08\text{ g/mol}}$

Value of $m$ is $\left(5.14-x\right)\text{ g}$ .

Value of $M$ is $153.33\text{ g/mol}$ .

Substitute the values in above equation for moles of $\text{CaO}$ .

  $\text{Moles of CaO}=\frac{\left(5.14-x\right)\text{ g}}{153.33\text{ g/mol}}$

Given reaction occurs as follows:

  $\begin{array}{l}\text{BaO}+{\text{CO}}_{\text{2}}\to {\text{BaCO}}_{\text{3}}\\ \text{CaO}+{\text{CO}}_{\text{2}}\to {\text{CaCO}}_{\text{3}}\end{array}$

Since total moles of $\text{BaO}$ and $\text{CaO}$ is equivalent to moles of ${\text{CO}}_{\text{2}}$ consumed, expression for moles of both is as follows:

  $\text{Moles of BaO}+\text{Moles of CaO}={\text{Moles of CO}}_2\text{ consumed}$

Moles of $\text{BaO}$ are $\left(\frac{x\text{ g}}{56.08\text{ g/mol}}\right)$ .

Moles of $\text{CaO}$ are $\left(\frac{\left(5.14-x\right)\text{ g}}{153.33\text{ g/mol}}\right)$ .

Moles of ${\text{CO}}_{\text{2}}$ consumed are $\text{0}\text{.0412 mol}$ .

Substitute the values in above equation.

  $\begin{array}{l}\text{Moles of BaO}+\text{Moles of CaO}={\text{Moles of CO}}_2\text{ consumed}\\ \left(\frac{x\text{ g}}{56.08\text{ g/mol}}\right)+\left(\frac{\left(5.14-x\right)\text{ g}}{153.33\text{ g/mol}}\right)=0.0412\text{ mol}\end{array}$

Value of $x$ is calculated as follows:

  $x=0.688\text{ g}$

Therefore mass of $\text{BaO}$ is $0.688\text{ g}$ .

Mass of $\text{CaO}$ is calculated as follows:

  $\begin{array}{c}\text{Mass of CaO}=\left(5.14-0.688\right)\text{ g}\\ =4.452\text{ g}\end{array}$

Expression for mass percent of substance is as follows:

  $\text{Mass perecnt of substance}=\left(\frac{\text{Mass of substance}}{\text{Mass of mixture}}\right)\left(100\text{ \%}\right)$

Mass of substance is $0.688\text{ g}$ .

Mass of mixture is $5.14\text{ g}$ .

Substitute the values in above formula for mass percent of $\text{BaO}$ .

  $\begin{array}{c}\text{Mass perecnt of BaO}=\left(\frac{\text{0}\text{.688 g}}{\text{5}\text{.14 g}}\right)\left(100\text{ \%}\right)\\ =13.38\text{ \%}\end{array}$

Mass of substance is $4.452\text{ g}$ .

Mass of mixture is $5.14\text{ g}$ .

Substitute the values in above formula for mass percent of $\text{CaO}$ .

  $\begin{array}{c}\text{Mass perecnt of CaO}=\left(\frac{4.452\text{ g}}{\text{5}\text{.14 g}}\right)\left(100\text{ \%}\right)\\ =86.61\text{ \%}\end{array}$

Hence, mass percent of $\text{BaO}$ is $13.38\text{ \%}$ and that of $\text{CaO}$ is $86.61\text{ \%}$ .

Conclusion

Mass percent of $\text{BaO}$ is $13.38\text{ \%}$ and that of $\text{CaO}$ is $86.61\text{ \%}$ .