Chapter 3, Problem 104AE

(a)

Interpretation Introduction

Interpretation: The explanation of the shape of the given graph is to be stated.

Concept introduction: Stoichiometric coefficients proportional to the respective chemical species can be correlated so as to compute their moles. The corresponding mass of these chemical species can then be worked out via their molar masses.The former concept is designated as stoichiometry. This concept can also be utilized to acquire limiting as well as excess reactants in given reaction.

Answer to Problem 104AE

The graph show increasing pattern as mass of $\text{NaCl}$ increases till fourth sample and becomes constant as ${\text{Cl}}_2$ gets completely consumed after fourth addition of $\text{Na}$ into the sample.

Explanation of Solution

The given graph is as follows.

  

Figure 1

The graph constitutes $7$ points for seven closed containers containing $7$ samples of $\text{NaCl}$ . All samples contain equal masses of ${\text{Cl}}_2$ gas and sodium is added $10.0\text{g}$ in the first sample, $20.0\text{g}$ in the second one followed up till $70.0\text{g}$ in seventh sample.

The graph shows an increasing pattern at first as the mass of $\text{Na}$ is increasing so the overall mass of $\text{NaCl}$ also increases. This increase is recorded till fourth sample. The graph becomes constant after fourth sample that is the mass of $\text{NaCl}$ does not show an increase. This indicates that after fourth addition, the amount of ${\text{Cl}}_2$ that is left would not be enough to react with $\text{Na}$ thereby showing a constant pattern till seventh sample.

Therefore, the graph show increasing pattern till fourth sample as mass of $\text{NaCl}$ increases but becomes constant as ${\text{Cl}}_2$ gets completely consumed after fourth addition of $\text{Na}$ into the sample.

(b)

Interpretation Introduction

Interpretation: The mass of $\text{NaCl}$ formed when $20.0\text{g}$ of $\text{Na}$ is used is to be calculated.

Concept introduction: Stoichiometric coefficients proportional to the respective chemical species can be correlated so as to compute their moles. The corresponding mass of these chemical species can then be worked out via their molar masses. The former concept is designated as stoichiometry. This concept can also be utilized to acquire limiting as well as excess reactants in given reaction.

Answer to Problem 104AE

The mass of $\text{NaCl}$ formed is $50.784\text{g}$ .

Explanation of Solution

The balanced reaction of $\text{Na}$ with ${\text{Cl}}_2$ is as follows.

  $2\text{Na}\left(s\right)+{\text{Cl}}_2\left(g\right)\to 2\text{NaCl}\left(s\right)$

The given mass of $\text{Na}$ is $20.0\text{g}$ .

The number of moles of $\text{Na}$ can be calculated as follows.

  $\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$  (1)

The molar mass of $\text{Na}$ is $23.0\text{g/mol}$ .

Substitute the values in equation (1).

  $\begin{array}{c}\text{Moles}\text{of}\text{Na}=\frac{20.0\text{g}}{23.0\text{g/mol}}\\ =0.869\text{mol}\end{array}$

The above reaction shows that $2$ mole of $\text{Na}$ forms $2$ mole of $\text{NaCl}$ . So, the number of moles of $\text{NaCl}$ formed will be same as that of $\text{Na}$ that is $0.869\text{mol}$ .

The mass of $\text{NaCl}$ formed can be calculated as follows.

  $\text{Mass}=\text{Moles}\times \text{Molar}\text{Mass}$  (2)

The molar mass of $\text{NaCl}$ is $58.44\text{g/mol}$ .

Substitute the values in equation (2).

  $\begin{array}{c}\text{Mass}\text{of}\text{NaCl}=0.869\text{mol}\times 58.44\text{g/mol}\\ =50.784\text{g}\end{array}$

Therefore, the mass of $\text{NaCl}$ formed is $50.784\text{g}$ .

(c)

Interpretation Introduction

Interpretation: The mass of ${\text{Cl}}_2$ in each container is to be calculated.

Concept introduction: Stoichiometric coefficients proportional to the respective chemical species can be correlated so as to compute their moles. The corresponding mass of these chemical species can then be worked out via their molar masses. The former concept is designated as stoichiometry. This concept can also be utilized to acquire limiting as well as excess reactants in given reaction.

Answer to Problem 104AE

The mass of ${\text{Cl}}_2$ in each container is $\text{61}\text{.7}\text{g}$ .

Explanation of Solution

The balanced reaction of $\text{Na}$ with ${\text{Cl}}_2$ is as follows.

  $2\text{Na}\left(s\right)+{\text{Cl}}_2\left(g\right)\to 2\text{NaCl}\left(s\right)$

The above reaction shows that $2$ mole of $\text{Na}$ reacts with $1$ mole of ${\text{Cl}}_2$ . The mass of $1$ mole of $\text{Na}$ is $23.0\text{g}$ . So the mass of $2$ mole of $\text{Na}$ will be $\left(2\times 23.0\right)\text{g}$ that is $46.0\text{g}$ .

The mass of $1$ mole of ${\text{Cl}}_2$ is $71\text{g}$ .

The graph shows that after fourth addition that is when $\text{Na}$ is $40\text{g}$ , ${\text{Cl}}_2$ gets used up completely. So, the mass of ${\text{Cl}}_2$ that will react with $40\text{g}$ of $\text{Na}$ will be present in each container.

The mass of ${\text{Cl}}_2$ in each container can be calculated as follows.

  $\begin{array}{c}\text{Mass}\text{of}{\text{Cl}}_2=\frac{71.0\text{g}{\text{Cl}}_2}{46.0\text{g}\text{Na}}\times 40.0\text{g}\text{Na}\\ =\text{61}\text{.7}\text{g}\end{array}$

Therefore, the mass of ${\text{Cl}}_2$ in each container is $\text{61}\text{.7}\text{g}$ .

(d)

Interpretation Introduction

Interpretation: The mass of $\text{NaCl}$ formed when $50.0\text{g}$ of $\text{Na}$ reacts is to be calculated.

Concept introduction: Stoichiometric coefficients proportional to the respective chemical species can be correlated so as to compute their moles. The corresponding mass of these chemical species can then be worked out via their molar masses. The former concept is designated as stoichiometry. This concept can also be utilized to acquire limiting as well as excess reactants in given reaction.

Answer to Problem 104AE

The mass of $\text{NaCl}$ formed is $101.7\text{g}$ .

Explanation of Solution

The balanced reaction of $\text{Na}$ with ${\text{Cl}}_2$ is as follows.

  $2\text{Na}\left(s\right)+{\text{Cl}}_2\left(g\right)\to 2\text{NaCl}\left(s\right)$

The given mass of $\text{Na}$ is $50.0\text{g}$ .

The molar mass of $\text{Na}$ is $23.0\text{g/mol}$ .

Since, the maximum amount of ${\text{Cl}}_2$ which reacts with $\text{Na}$ is when $\text{Na}$ is $40.0\text{g}$ . The mass of $\text{Na}$ that actually reacts will be $40.0\text{g}$ .

The number of moles of $\text{Na}$ can be calculated using equation (1) as follows.

  $\begin{array}{c}\text{Moles}\text{of}\text{Na}=\frac{40.0\text{g}}{23.0\text{g/mol}}\\ =1.74\text{mol}\end{array}$

The above reaction shows that $2$ mole of $\text{Na}$ forms $2$ mole of $\text{NaCl}$ . So, the number of moles of $\text{NaCl}$ formed will be same as that of $\text{Na}$ that is $1.74\text{mol}$ .

The molar mass of $\text{NaCl}$ is $58.44\text{g/mol}$ .

Substitute the values in equation (2) to calculate the mass of $\text{Na}$ as follows.

  $\begin{array}{c}\text{Mass}\text{of}\text{NaCl}=1.74\text{mol}\times 58.44\text{g/mol}\\ =101.7\text{g}\end{array}$

Therefore, the mass of $\text{NaCl}$ formed is $101.7\text{g}$ .

(e)

Interpretation Introduction

Interpretation: The left over reactants and their masses in parts b and d are to be calculated.

Concept introduction: Stoichiometric coefficients proportional to the respective chemical species can be correlated so as to compute their moles. The corresponding mass of these chemical species can then be worked out via their molar masses. The former concept is designated as stoichiometry. This concept can also be utilized to acquire limiting as well as excess reactants in given reaction.

Answer to Problem 104AE

The leftover reagent in part b is ${\text{Cl}}_2$ and has a mass of $30.83\text{g}$ and in part d is $\text{Na}$ having mass of $\text{10}\text{.0}\text{g}$ .

Explanation of Solution

The balanced reaction of $\text{Na}$ with ${\text{Cl}}_2$ is as follows.

  $2\text{Na}\left(s\right)+{\text{Cl}}_2\left(g\right)\to 2\text{NaCl}\left(s\right)$

The above reaction shows that $2$ mole of $\text{Na}$ reacts with $1$ mole of ${\text{Cl}}_2$. The mass of $1$ mole of $\text{Na}$ is $23.0\text{g}$ . So the mass of $2$ mole of $\text{Na}$ will be $\left(2\times 23.0\right)\text{g}$ that is $46.0\text{g}$ .

The mass of $1$ mole of ${\text{Cl}}_2$ is $71\text{g}$ .

The graph shows that ${\text{Cl}}_2$ is in excess till fourth addition of $\text{Na}$ . So ${\text{Cl}}_2$ will be the leftover reactant in part b.

The given mass of $\text{Na}$ in part b is $20.0\text{g}$ .

The mass of ${\text{Cl}}_2$ which will react will $20.0\text{g}$ of $\text{Na}$ can be calculated as follows.

  $\begin{array}{c}\text{Mass}\text{of}{\text{Cl}}_2=\frac{71.0\text{g}{\text{Cl}}_2}{46.0\text{g}\text{Na}}\times 20.0\text{g}\text{Na}\\ =30.87\text{g}\end{array}$

The mass of ${\text{Cl}}_2$ that will be leftover in part b can be calculated as follows.

  $\begin{array}{c}\text{Mass}\text{of}{\text{Cl}}_2\text{left}=\text{Mass}\text{of}{\text{Cl}}_2\text{in}\text{each}\text{container}\\ -\text{Mass}\text{of}{\text{Cl}}_2\text{reacts}\end{array}$  (3)

Substitute the values in equation (3).

  $\begin{array}{c}\text{Mass}\text{of}{\text{Cl}}_2\text{left}=\text{61}\text{.7}\text{g}-30.87\text{g}\\ =30.83\text{g}\end{array}$

The graph shows that in part d, ${\text{Cl}}_2$ gets consumed, so $\text{Na}$ will be the leftover reagent.

The mass of $\text{Na}$ in part d is $50.0\text{g}$ out of which only $40.0\text{g}$ will react. So, the left over mass of $\text{Na}$ can be calculated as follows.

  $\text{Mass}\text{of}\text{Na}\text{left}=\text{Mass}\text{of}\text{Na}\text{in}\text{part}\text{d}-\text{Mass}\text{of}\text{Na}\text{reacting}$  (4)

Substitute the values in equation (4).

  $\begin{array}{c}\text{Mass}\text{of}\text{Na}\text{left}=50.0\text{g}-40.0\text{g}\\ =\text{10}\text{.0}\text{g}\end{array}$

Therefore, the leftover reagent in part b is ${\text{Cl}}_2$ and has a mass of $30.83\text{g}$ and in part d is $\text{Na}$ having mass of $\text{10}\text{.0}\text{g}$ .