Chapter 3, Problem 160AE

You have seven closed containers, each with equal masses of chlorine gas (Cl₂) . You add 10.0 g of sodium to the first sample, 20.0 g of sodium to the second sample, and so on (adding 70.0 g of sodium to the seventh sample). Sodium and chlorine react to form sodium chloride according to the equation

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

After each reaction is complete, you collect and measure the amount of sodium chloride formed. A graph of your results is shown below.

Answer the following questions:

a. Explain the shape of the graph.

b. Calculate the mass of NaCl formed when 20.0 g of sodium is used.

c. Calculate the mass of Cl₂ in each container.

d. Calculate the mass of NaCl formed when 50.0 g of sodium is used.

e. Identify the leftover reactant, and determine its mass for parts b and d above.

Interpretation Introduction

Interpretation: Seven closed containers, having equal amount of chlorine gas present in them, have been given. Sodium is added to the containers in the given pattern. After the completion of the reaction between sodium and chlorine, the amount of sodium chloride formed is measured. The stated questions related to the given process are to be answered.

Concept introduction: The amount of the sodium chloride formed depends upon the amount of sodium and chlorine gas present in each of the given containers.

To determine: The explanation regarding the shape of the graph.

Explanation of Solution

The amount of sodium chloride formed increases on increasing the quantity of sodium from the container one to four.

In the fifth container the addition of a greater amount of sodium than added to the previous container did not increase the amount of sodium chloride formed.

This indicates that the amount of chlorine gas present in the respective container is not sufficient to react with amount of sodium been added to the container. Hence, the amount of sodium chloride formed in the subsequent containers has a constant value.

Conclusion

The amount of chlorine gas present in the containers five, six and seven reacts with an equal amount of sodium to form an equal amount of sodium chloride.

Interpretation Introduction

Interpretation: Seven closed containers, having equal amount of chlorine gas present in them, have been given. Sodium is added to the containers in the given pattern. After the completion of the reaction between sodium and chlorine, the amount of sodium chloride formed is measured. The stated questions related to the given process are to be answered.

Concept introduction: The amount of the sodium chloride formed depends upon the amount of sodium and chlorine gas present in each of the given containers.

To determine: The mass of $\text{NaCl}$ formed on using $20.0\text{g}$ of sodium.

Explanation of Solution

Given

The stated reaction is,

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

The atomic mass of sodium is $23\text{g}$.

The molar mass of $\text{NaCl}$ $\begin{array}{l}=\text{Na}+\text{Cl}\\ =\left(23+35.5\right)\text{g}\\ =58.5\text{g}\end{array}$

According to the given reaction,

$2\text{mol}$ of $\text{Na}$ produce $\text{NaCl}$ $=2\text{mol}$

$1\text{mol}$ of $\text{Na}$ produces $\text{NaCl}$ $=1\text{mol}$

$23\text{g}$ of $\text{Na}$ produce $\text{NaCl}$ $=58.5\text{g}$

$1\text{g}$ of $\text{Na}$ produce $\text{NaCl}$ $=\left(\frac{58.5}{23}\right)\text{g}$

$20.0\text{g}$ of $\text{Na}$ produce $\text{NaCl}$ $\begin{array}{l}=\left(\frac{58.5\times 20.0}{23}\right)\text{g}\\ \approx \underset{\_}{51g}\end{array}$

Conclusion

The mass of $\text{NaCl}$ formed on using $20.0\text{g}$ of sodium is $\underset{\_}{51g}$.

Interpretation Introduction

Interpretation: Seven closed containers, having equal amount of chlorine gas present in them, have been given. Sodium is added to the containers in the given pattern. After the completion of the reaction between sodium and chlorine, the amount of sodium chloride formed is measured. The stated questions related to the given process are to be answered.

Concept introduction: The amount of the sodium chloride formed depends upon the amount of sodium and chlorine gas present in each of the given containers.

To determine: The mass of ${\text{Cl}}_2$ present in each container.

Explanation of Solution

Given

The stated reaction is,

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

The atomic mass of sodium is $23\text{g}$.

The molar mass of $\text{NaCl}$ $\begin{array}{l}=\text{Na}+\text{Cl}\\ =\left(23+35.5\right)\text{g}\\ =58.5\text{g}\end{array}$

According to the given graph, chlorine gas present reacts completely with $40\text{g}$ of sodium.

According to the given reaction,

$2\text{mol}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $=1\text{mol}$

$\left(2\times 23\right)\text{g}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $=\left(2\times 35.5\right)\text{g}$

$1\text{g}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $=\left(\frac{2\times 35.5}{\left(2\times 23\right)}\right)\text{g}$

$40\text{g}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $\begin{array}{l}=\left(\frac{2\times 35.5\times 40}{\left(2\times 23\right)}\right)\text{g}\\ =\underset{\_}{61.7g}\end{array}$

Conclusion

The mass of ${\text{Cl}}_2$ present in each container is $\underset{\_}{61.7g}$.

Interpretation Introduction

Interpretation: Seven closed containers, having equal amount of chlorine gas present in them, have been given. Sodium is added to the containers in the given pattern. After the completion of the reaction between sodium and chlorine, the amount of sodium chloride formed is measured. The stated questions related to the given process are to be answered.

Concept introduction: The amount of the sodium chloride formed depends upon the amount of sodium and chlorine gas present in each of the given containers.

To determine: The mass of $\text{NaCl}$ formed on using $50.0\text{g}$ of sodium.

Explanation of Solution

Given

The stated reaction is,

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

The atomic mass of sodium is $23\text{g}$.

The molar mass of $\text{NaCl}$ $\begin{array}{l}=\text{Na}+\text{Cl}\\ =\left(23+35.5\right)\text{g}\\ =58.5\text{g}\end{array}$

According to the given graph, chlorine gas present reacts completely with $40\text{g}$ of sodium.

Hence, only $40.0\text{g}$ of sodium will be used in the stated reaction.

According to the given reaction,

$2\text{mol}$ of $\text{Na}$ produce $\text{NaCl}$ $=2\text{mol}$

$1\text{mol}$ of $\text{Na}$ produces $\text{NaCl}$ $=1\text{mol}$

$23\text{g}$ of $\text{Na}$ produce $\text{NaCl}$ $=58.5\text{g}$

$1\text{g}$ of $\text{Na}$ produce $\text{NaCl}$ $=\left(\frac{58.5}{23}\right)\text{g}$

$40.0\text{g}$ of $\text{Na}$ produce $\text{NaCl}$ $\begin{array}{l}=\left(\frac{58.5\times 40.0}{23}\right)\text{g}\\ \approx \underset{\_}{101.7g}\end{array}$

Conclusion

The mass of $\text{NaCl}$ formed on using $20.0\text{g}$ of sodium is $\underset{\_}{101.7g}$.

Interpretation Introduction

Interpretation: Seven closed containers, having equal amount of chlorine gas present in them, have been given. Sodium is added to the containers in the given pattern. After the completion of the reaction between sodium and chlorine, the amount of sodium chloride formed is measured. The stated questions related to the given process are to be answered.

Concept introduction: The amount of the sodium chloride formed depends upon the amount of sodium and chlorine gas present in each of the given containers.

To determine: The left over reactant in parts (b) and (d) and their respective mass.

Explanation of Solution

The leftover reactant in part (b) is the chlorine gas. The mass of chlorine left unreacted is $\underset{\_}{30.8g}$. The leftover reactant in part (d) is sodium. The mass of sodium left unreacted is $\underset{\_}{10.0g}$.

To determine: The left over reactant in parts (b) and its respective mass.

The leftover reactant in part (b) is the chlorine gas. The mass of chlorine left unreacted is $\underset{\_}{30.8g}$.

Given

The stated reaction is,

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

The atomic mass of sodium is $23\text{g}$.

The molar mass of $\text{NaCl}$ $\begin{array}{l}=\text{Na}+\text{Cl}\\ =\left(23+35.5\right)\text{g}\\ =58.5\text{g}\end{array}$

According to the given reaction,

$2\text{mol}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $=1\text{mol}$

$\left(2\times 23\right)\text{g}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $=\left(2\times 35.5\right)\text{g}$

$1\text{g}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $=\left(\frac{2\times 35.5}{\left(2\times 23\right)}\right)\text{g}$

$20\text{g}$ of $\text{Na}$ reacts with ${\text{Cl}}_2$ $\begin{array}{l}=\left(\frac{2\times 35.5\times 20}{\left(2\times 23\right)}\right)\text{g}\\ =30.9\text{g}\end{array}$

The mass of chlorine present in the container is $\text{61}\text{.7}\text{g}$.

The amount of chlorine left unreacted is $\begin{array}{l}=\left(61.7-30.9\right)\text{g}\\ =\underset{\_}{30.8g}\end{array}$

To determine: The left over reactant in parts (d) and its respective mass.

The leftover reactant in part (d) is sodium. The mass of sodium left unreacted is $\underset{\_}{10.0g}$.

According to the given graph, chlorine gas present reacts completely with $40\text{g}$ of sodium.

Hence, only $40.0\text{g}$ of sodium will be used in the stated reaction.

The amount of sodium left unreacted is $\begin{array}{l}=\left(50.0-40.0\right)\text{g}\\ =\underset{\_}{10.0g}\end{array}$

Conclusion

The mass of chlorine gas left unreacted in the stated reaction $\underset{\_}{30.8g}$.

The mass of sodium left unreacted in the stated reaction $\underset{\_}{10.0g}$