Chapter 10, Problem 111AP

A mixture of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ and MgCO}}_{\text{3}}$ of mass 7.63 g is combined with an excess of hydrochloric acid. The ${\text{CO}}_{\text{2}}$ gas generated occupies a volume of 1.67 L at 1.24 atm and $$. From these data, calculate the percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture.

Interpretation Introduction

Interpretation:

The percent by mass of ${\text{Na}}_2{\text{CO}}_3$ in the sample is to be calculated.

Concept Introduction:

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

$PV=nRT.$

Here, $n$ is number of moles, $R$ is universal gas constant, $T$ is absolute temperature, $V$ is volume, and $P$ is pressure.

The formula for conversion of temperature from Celsius to kelvin is represented as

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

Answer to Problem 111AP

Solution: $33.1\%$.

Explanation of Solution

Given information:

Mass $m=7.63\text{ g}$

Volume $V=1.67\text{ L}$

Temperature $T=26°\text{C}$

Pressure $P=1.24\text{ atm}$

The reaction of ${\text{Na}}_{\text{2}}{\text{CO}}_3$ with hydrochloric acid is as follows:

${\text{Na}}_2{\text{CO}}_3\left(s\right)+2\text{ HCl}\left(aq\right)\to 2\text{}\text{ NaCl}\left(aq\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right).$

The reaction of $\text{Mg}{\text{CO}}_3$ with hydrochloric acid is

${\text{MgCO}}_3\left(s\right)+2\text{ HCl}\left(aq\right)\to {\text{MgCl}}_2\left(aq\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right).$

The temperature is $26°\text{C}\text{.}$

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

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(26+273.15\right)\\ =299.15\text{ K}\text{.}\end{array}$

The equation for an ideal gas is

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

Calculate the number of moles of carbon dioxide as follows:

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

Substitute $1.67\text{ L}$ for $V$, $299.15\text{ K}$ for $T$, $1.24\text{ atm}$ for $P,$ and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation.

$\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{\left(1.24\cancel{\text{atm}}\right)\left(1.67\cancel{\text{L}}\right)}{\left(0.08206\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(299.15\cancel{\text{K}}\right)}\\ =\frac{2.07}{24.54}\text{ mol}\\ =0.0844\text{ mol}\end{array}$

From the above balanced stoichiometric equation it is seen that the ratio of number of moles of carbon dioxide and ${\text{Na}}_2{\text{CO}}_3$ or ${\text{MgCO}}_3$ is $1:1$.

So, the number of moles of the mixture is

$n_{{\text{Na}}_2{\text{CO}}_3}+n_{{\text{MgCO}}_3}=0.0844\text{ mol}\text{.}$

The mass of the sample is $7.63\text{ g}\text{.}$

Let the mass of ${\text{Na}}_2{\text{CO}}_3$ be $x$.

So, the mass of ${\text{MgCO}}_3$ will be $7.63-x$.

The molar mass of ${\text{Na}}_2{\text{CO}}_3$ is $106\text{ g}/\text{mol}\text{.}$

The molar mass of ${\text{MgCO}}_3$ is $84.3\text{ g}/\text{mol}\text{.}$

The number of moles can be calculated as

$n=\frac{m}{M},$

where m is the given mass and M is the molar mass

So, the moles of sample can be written as

$\frac{m_{{\text{Na}}_2{\text{CO}}_3}}{M_{{\text{Na}}_2{\text{CO}}_3}}+\frac{m_{{\text{MgCO}}_3}}{M_{{\text{MgCO}}_3}}=0.0844\text{ mol}\text{.}$

Substitute $106\text{ g}/\text{mol}$ for $M_{{\text{Na}}_2{\text{CO}}_3}$,$84.3\text{ g}/\text{mol}$ for $M_{{\text{MgCO}}_3}$, $x$ for $m_{{\text{Na}}_2{\text{CO}}_3},$ and $7.63-x$ for $m_{{\text{MgCO}}_3}$ in the above equation.

$\begin{array}{c}\frac{x}{106\text{ g}/\text{mol}}+\frac{\left(7.63-x\right)}{84.3\text{ g}/\text{mol}}=0.0844\text{ mol}\\ \frac{\left(84.3x\right)+106\left(7.63-x\right)}{\left(106\text{ g}/\text{mol}\right)\left(84.3\text{ g}/\text{mol}\right)}=0.0844\text{ mol}\\ \frac{84.3x-106x+808.78}{8935.8\text{ g}/\text{mol}}=0.0844\text{ mol}\end{array}$

Rearrange the above equation as follows:

$\begin{array}{c}x=\frac{54.598}{21.7}\\ =2.52\text{ g}\text{.}\end{array}$

So, the mass of ${\text{Na}}_2{\text{CO}}_3$ is $2.52\text{ g}\text{.}$

Now, the percent by mass of ${\text{Na}}_2{\text{CO}}_3$ in the sample of $7.63\text{ g}$ is

$\begin{array}{c}{\text{\%Na}}_2{\text{CO}}_3=\frac{2.52\cancel{\text{g}}}{7.63\cancel{\text{g}}}\times 100\%\\ =0.331\times 100\%\\ =33.12\%\end{array}$

Conclusion

The percent by mass of ${\text{Na}}_2{\text{CO}}_3$ in the sample of $7.63\text{ g}$ is $33.1\%$.