Chapter 10, Problem 169AP

Acidic oxides such as carbon dioxide react with basic oxides like calcium oxide $\left(\text{CaO}\right)$ and barium oxide $\left(\text{BaO}\right)$ to form salts (metal carbonates). (a) Write equations representing these two reactions. (b) A student placed a mixture of BaO and CaO of combined mass 4.88 g in a 1.46-L flask containing carbon dioxide gas at $\text{35ºC}$ and 746 mmHg. After the reactions were complete, she found that the ${\text{CO}}_2$ pressure had dropped to 252 mmHg. Calculate the percent composition by mass of the mixture. Assume that the volumes of the solids are negligible.

Interpretation Introduction

Interpretation:

The balanced equations representing the reactions and the percent composition by mass of each reactant in the sample are to be determined with given mass and volume of carbon dioxide, pressure, and volume.

Concept Introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by

$P=R\frac{nT}{V}.$

A balanced chemical equation follows the law of conservation of mass, according to which the number of different atoms in reactant side and product side must be equal.

The number of moles can be calculated as:

$m=\frac{wt}{M}.$

Here, m is the number of moles, wt is the given mass of compound, and M is the molar mass of the compound.

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

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

The relationship between $\text{atm}$ and $\text{mm Hg}$ can be expressed as:

$1\text{ mm Hg}=\frac{1}{760}\text{ atm}\text{.}$

To convert milliliters into liters, the conversion factor is $\left(\frac{\text{1 atm}}{\text{760 mm Hg}}\right).$

Answer to Problem 169AP

Solution:

(a)

The balanced chemical equations for the two reactions are as:

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

$\text{BaO}\left(s\right)+{\text{CO}}_2\left(g\right) \to {\text{BaCO}}_3\left(s\right).$

(b)

Percent by mass of $\text{BaO}$ in the sample of $4.88\text{ g}$ is $89.5\%$ and of $\text{CaO}$ is $10.5\%$.

Explanation of Solution

a)Equations representing given two reactions

The balanced reaction of calcium oxide with carbon dioxide is as

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

The balanced reaction of barium oxide with carbon dioxide is as

$\text{BaO}\left(s\right)+{\text{CO}}_2\left(g\right) \to {\text{BaCO}}_3\left(s\right).$

b) The percent composition by mass of the mixture

The combined mass for mixture of BaO and CaOis $4.88\text{ g}$.

Both the salts are mixed in a flask of volume $1.46\text{ L}$ containing ${\text{CO}}_2$ at a pressure $746\text{ mmHg}$.

The equation for an ideal gas is as

$PV=nRT.$

For ${\text{CO}}_2$, this can be rewritten as

$n_{{\text{CO}}_2}=\frac{P{V}_{{\text{CO}}_2}}{RT}.$

The conversion of temperature from degree Celsius to Kelvin can be done as follows:

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

The absolute value of $T$ is as:

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

The pressure is $746\text{ mmHg}$.

Convert mmHg to atm as:

$\begin{array}{c}746\text{ mmHg}=\left(746\cancel{\text{mmHg}}\right)\left(\frac{1\text{ atm}}{760\cancel{\text{mmHg}}}\right)\\ =0.982\text{ atm}\text{.}\end{array}$

Substitute the values $1.46\text{ L}$ for $V$, $308.15\text{ K}$ for $T$, $0.982\text{ atm}$ for $P$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the ideal gas equation as follows:

$\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{\left(0.982\cancel{\text{atm}}\right)\left(1.46\cancel{\text{L}}\right)}{\left(0.08206\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(308.15\cancel{\text{K}}\right)}\\ =\frac{1.43}{25.82}\text{ mol}\\ =0.0567\text{ mol}\text{.}\end{array}$

After the completion of reaction:

The pressure of ${\text{CO}}_2$ is $252\text{ mmHg}$.

Convert mmHg to atm as:

$\begin{array}{c}252\text{ mmHg}=\left(252\cancel{\text{mmHg}}\right)\left(\frac{1\text{ atm}}{760\cancel{\text{mmHg}}}\right)\\ =0.33\text{ atm}\text{.}\end{array}$

Substitute the values $1.46\text{ L}$ for $V$, $308.15\text{ K}$ for $T$, $0.33\text{ atm}$ for $P$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation as follows:

$\begin{array}{c}{n}_{{\text{CO}}_2\text{'}}=\frac{\left(0.33\cancel{\text{atm}}\right)\left(1.46\cancel{\text{L}}\right)}{\left(0.08206\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(308.15\cancel{\text{K}}\right)}\\ =\frac{0.482}{25.82}\text{ mol}\\ =0.019\text{ mol}\text{.}\end{array}$

So, the quantity of carbon dioxide spent in the reaction is determined as

$\begin{array}{c}{n}_{{\text{CO}}_2}-n_{{\text{CO}}_2\text{'}}=0.0567\text{ mol}-0.0191\text{ mol}\\ =0.0376\text{ mol}\text{.}\end{array}$

The ratio of number of moles of carbon dioxide and $\text{BaO}$ or $\text{CaO}$ is $1:1$.

So, the number of moles of the mixture is determined as

$n_{\text{BaO}}+n_{\text{CaO}}=0.0376\text{ mol}\text{.}$

The mass of the sample is $4.88\text{ g}$.

Consider the mass of $\text{BaO}$ to be $x$.

So, the mass of $\text{CaO}$ will be $\left(4.88-x\right)\text{ g}$.

The molar mass of $\text{BaO}$, $M_{\text{BaO}}=153.3\text{ g}/\text{mol}\text{.}$

The molar mass of $\text{CaO}$, $M_{\text{CaO}}=56.08\text{ g}/\text{mol}\text{.}$

The number of moles can be calculated as

$m=\frac{wt}{M}.$

Here, m is the number of moles, wt is the given mass of compound, and M is the molar mass of the compound.

So, the moles of sample are as

$\frac{m_{\text{BaO}}}{M_{\text{BaO}}}+\frac{m_{\text{CaO}}}{M_{\text{CaO}}}=0.0376\text{ mol}\text{.}$

Substitute the values $x$ for $m_{\text{BaO}}$, $x-4.88\text{ g}$ for $m_{\text{CaO}}$, $153.3\text{ g}/\text{mol}$ for $M_{\text{BaO}}$, and $56.08\text{ g}/\text{mol}$ for $M_{\text{CaO}}$ in the above equation as follows:

$\begin{array}{c}\frac{x}{153.3\text{ g}/\text{mol}}+\frac{\left(4.88-x\right)}{56.08\text{ g}/\text{mol}}=0.0376\text{ mol}\\ \frac{\left(56.08x\right)+153.3\left(4.88-x\right)}{\left(153.3\text{ g}/\text{mol}\right)\left(56.08\text{ g}/\text{mol}\right)}=0.0376\text{ mol}\\ \frac{56.08x-153.3x+748.1}{8597.06\text{ g}/\text{mol}}=0.0376\text{ mol}\end{array}$

This can be rearranged as

$\begin{array}{c}56.08x-153.3x+748.1=0.0376\text{ mol}\times 8597.06\text{ g}/\text{mol}\\ 748.1-\text{97}\text{.22}x=323.24\\ \text{97}\text{.22}x=748.1-323.24\\ =424.86\end{array}$

So, the mass of $\text{BaO}$ is

$\begin{array}{c}x=\frac{424.86}{97.22}\\ =4.37\text{ g}\text{.}\end{array}$

The mass of $\text{CaO}$ is

$\begin{array}{l}\left(4.88-4.37\right)\text{ g}\\ =0.513\text{ g}\text{.}\end{array}$

Now, the percent by mass of $\text{BaO}$ and $\text{CaO}$ in the sample of $4.88\text{ g}$ is

For $\text{BaO}$

$\begin{array}{c}\text{\%BaO}=\frac{4.37\text{ g}}{4.88\text{ g}}\times 100\%\\ =0.895\times 100\%\\ =89.5\%.\end{array}$

For $\text{CaO}$

$\begin{array}{c}\text{\%CaO}=\frac{0.513\text{ g}}{4.88\text{ g}}\times 100\%\\ =0.105\times 100\%\\ =10.5\%.\end{array}$

Hence, the percent by mass of $\text{BaO}$ in the sample of $4.88\text{ g}$ is $89.5\%$ and of $\text{CaO}$ is $10.5\%$.