Chapter 10, Problem 102IL

Interpretation Introduction

Interpretation:

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{NaHCO}}_{\text{3}}$ in the mixture should be calculated.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\\ \text{where,}\\ \text{n = moles of gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties. At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Answer to Problem 102IL

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture is $\text{57}\text{.6\%}$ and the weight percentage for ${\text{NaHCO}}_{\text{3}}$ in the mixture is $\text{42}\text{.4\%}$.

Given:

  $\begin{array}{l}\text{Total amount of the mixture = 2}\text{.50 g}\\ {\text{Volume of CO}}_{\text{2}}\text{=}\text{665mL}\text{=}{\text{665×10}}^{\text{-3}}\text{L}\\ {\text{Pressure of CO}}_{\text{2}}\text{=}\text{735mmHg}\text{=}\frac{\text{735mmHg}}{\text{760mmHg}}=0.967atm\\ {\text{Temperature of CO}}_{\text{2}}{\text{= 25}}^{\text{o}}\text{C = 273}\text{.15K+25}\text{=}\text{289}\text{.15K}\end{array}$

Explanation of Solution

The balanced chemical equation for the reaction of the two given gases with $HCl$ is as follows,

  $\begin{array}{l}NaHC{O}_3+HCl\to NaCl+H_{2}O+C{O}_2\\ N{a}_{2}C{O}_3+HCl\to 2NaCl+H_{2}O+C{O}_2\end{array}$

First using the ideal gas equation the moles of ${\text{CO}}_{\text{2}}$ produced is calculated as follows,

Here, we carry an extra significant figure throughout this calculation to limit rounding errors.

  $\begin{array}{l}{\text{ n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{(0}{\text{.967atm)(665×10}}^{\text{-3}}\text{L)}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(289}\text{.15)}\text{K}}\text{=}\text{0}\text{.02709}\text{mol}{\text{CO}}_{\text{2}}\end{array}$

From the balanced chemical equation it is clear that one mole of both the gases gives rise to one mole of ${\text{CO}}_{\text{2}}$ hence, the total calculated amount of $\text{0}\text{.02709}\text{mol}{\text{CO}}_{\text{2}}$ should be then produced by $\text{0}\text{.02709}\text{mol}$ of gases present in the given mixture.

Hence, there exists 1:1 mole ratio between ${\text{CO}}_{\text{2}}$ and both reactants $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3 }}{\text{and MgCO}}_{\text{3}}\right)\text{,}$ which show that $\text{0}\text{.02709}$ of the mixture must have reacted.

$\text{mol}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}\text{mol}{\text{NaHCO}}_{\text{3}}\text{=}\text{0}\text{.02709}\text{mol}$

${\text{Let x be the mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ in the mixture, then (2}{\text{.50- x) is the mass of NaHCO}}_{\text{3 }}\text{in the mixture}\text{.}$    $\begin{array}{l}\left(\text{x}\text{g}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{105}\text{.99}\text{g}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\text{+}\left[\text{(2}\text{.50-x)g}{\text{MgCO}}_{\text{3}}\text{×}\frac{\text{1}{\text{molNaHCO}}_{\text{3}}}{\text{84}\text{.007}\text{g}{\text{NaHCO}}_{\text{3}}}\right]\\ \text{=}\text{0}\text{.08436}\text{mol}\end{array}$

  $\begin{array}{l}\text{9}{\text{.4349×10}}^{\text{-3}}\text{x+}\left(\text{2}\text{.50-x}\right)\text{0}\text{.012 = 0}\text{.02709}\\ \text{9}{\text{.435×10}}^{\text{-3}}\text{x+0}\text{.03-0}\text{.012x = 0}\text{.02709}\\ \cancel{\text{-}}\text{2}\text{.565×}\cancel{{\text{10}}^{\text{-3}}}\text{x = }\cancel{\text{-}}\text{3}\text{.7×}\cancel{{\text{10}}^{\text{-3}}}\\ \text{x = 1}\text{.44g}\\ \text{x = 1}{\text{.44 g = mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ in the mixture}\end{array}$

The mass of the remaining ${\text{NaHCO}}_{\text{3}}$ in the mixture is determined by subtracting the amount of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ from the total amount of the mixture as follows gives the mass of the other gas present in the mixture.

  $\begin{array}{l}2.50g-1.44g=1.66g\\ \end{array}$

Determine the percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture as follows,

  $\begin{array}{l}\text{mass\%}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{=}\frac{\text{mass}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{mass}\text{of}\text{mixture}}\text{×}\text{100\%}\\ \text{=}\frac{\text{1}\text{.44 g}}{\text{2}\text{.50 g}}\text{×}\text{100\%}=57.6\%\end{array}$

Similarly, the weight percentage for the other gas in the mixture is as follows,

  $\begin{array}{l}\text{mass\%}{\text{NaHCO}}_{\text{3}}\text{=}\frac{\text{mass}{\text{NaHCO}}_{\text{3}}}{\text{mass}\text{of}\text{mixture}}\text{×}\text{100\%}\\ \text{=}\frac{\text{1}\text{.06 g}}{\text{2}\text{.50 g}}\text{×}\text{100\%}=42.4\%\end{array}$

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture is calculated using the ratio between $\text{mass}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and mass of mixture.  The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture is

  $\text{33}\text{.1\%}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

Conclusion

The percent composition by mass of two given gases in the mixture was calculated by using the ideal gas equation and the expression used for moles calculation.