Chapter 4, Problem 4.115P

(a)

Interpretation Introduction

Interpretation:

The reactant present in excess when $1.62\text{ g}$ of lithium is mixed with $6.50\text{ g}$ of oxygen is to be identified.

Concept introduction:

The redox reaction can be classified into three types depending upon the number of reactants and products as follows:

1. Combination redox reaction

2. Decomposition redox reaction

3. Displacement redox reactions

Combination redox reactions are the reactions in which two or more reactants combine to form a single product. In displacement redox reactions, substances on both sides of the equation remain the same but the atoms exchange places in order to form the product while in decomposition reaction, one compound decompose to form one or more product.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

Answer to Problem 4.115P

The reactant present in excess when $1.62\text{ g}$ of lithium is mixed with $6.50\text{ g}$ of oxygen is ${\text{O}}_2$.

Explanation of Solution

Lithium combines with oxygen molecule to form lithium oxide $\left({\text{Li}}_{\text{2}}\text{O}\right)$. The balanced chemical equation of the redox reaction is:

$\text{4Li}\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{Li}}_2\text{O}\left(s\right)$

Four moles of $\text{Li}$ combine with one mole of ${\text{O}}_2$ to give two mole of ${\text{Li}}_{\text{2}}\text{O}$.

The molecular mass of $\text{Li}$ is $6.941\text{g/mol}$.

The formula to calculate moles of ${\text{Li}}_{\text{2}}\text{O}$ when $\text{Li}$ is limiting reagent is:

$\text{Moles}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\frac{\text{mass of}\text{Li}\left(\text{g}\right)}{\text{molecular mass}\text{of}\text{Li}\left(\text{g/mol}\right)}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{4\text{mol}\text{Li}}\right)\right]$                        (1)

Substitute $1.62\text{ g}$ for mass of $\text{Li}$ and $6.941\text{g/mol}$ for molecular mass of $\text{Li}$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\frac{1.62\text{ g}}{6.941\text{g/mol}}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{4\text{mol}\text{Li}}\right)\right]\\ =\left[\left(0.23339\text{mol}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{4\text{mol}\text{Li}}\right)\right]\\ =0.1166979\text{mol}\end{array}$

The molecular mass of ${\text{O}}_2$ is $32.00\text{g/mol}$.

The formula to calculate moles of ${\text{Li}}_{\text{2}}\text{O}$ when ${\text{O}}_2$ is limiting reagent is:

$\text{Moles}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\frac{\text{mass of}{\text{O}}_2\left(\text{g}\right)}{\text{molecular mass}\text{of}{\text{O}}_2\left(\text{g/mol}\right)}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{1\text{mol}{\text{O}}_2}\right)\right]$                       (2)

Substitute $6.50\text{ g}$ for mass of ${\text{O}}_2$ and $32.00\text{g/mol}$ for molecular mass of ${\text{O}}_2$ in the equation (2).

$\begin{array}{c}\text{Moles}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\frac{6.50\text{ g}}{32.00\text{g/mol}}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{1\text{mol}{\text{O}}_2}\right)\right]\\ =\left[\left(0.203125\text{mol}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{1\text{mol}{\text{O}}_2}\right)\right]\\ =0.40625\text{mol}\end{array}$

$\text{Li}$ is limiting reagent in the reaction as the moles of ${\text{Li}}_{\text{2}}\text{O}$ produced is less in this case as compared to when ${\text{O}}_2$ is the limiting agent.

The reactant present in excess concentration in reaction is ${\text{O}}_2$.

Conclusion

The reactant present in excess when $1.62\text{ g}$ of lithium is mixed with $6.50\text{ g}$ of oxygen is ${\text{O}}_2$.

(b)

Interpretation Introduction

Interpretation:

The moles of product formed when $1.62\text{ g}$ of lithium is mixed with $6.50\text{ g}$ of oxygen is to be calculated.

Concept introduction:

The redox reaction can be classified into three types depending upon the number of reactants and products as follows:

1. Combination redox reaction

2. Decomposition redox reaction

3. Displacement redox reactions

Combination redox reactions are the reactions in which two or more reactants combine to form a single product. In displacement redox reactions, substances on both sides of the equation remain the same but the atoms exchange places in order to form the product while in decomposition reaction, one compound decompose to form one or more product.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

Answer to Problem 4.115P

The moles of product formed when $1.62\text{ g}$ of lithium is mixed with $6.50\text{ g}$ of oxygen is $0.1166979\text{mol}$.

Explanation of Solution

Lithium combines with oxygen molecule to form lithium oxide $\left({\text{Li}}_{\text{2}}\text{O}\right)$. The balanced chemical equation of the redox reaction is:

$\text{4Li}\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{Li}}_2\text{O}\left(s\right)$

Lithium is the limiting agent in the reaction.

Four moles of $\text{Li}$ combine with one mole of ${\text{O}}_2$ to give two mole of ${\text{Li}}_{\text{2}}\text{O}$.

The molecular mass of $\text{Li}$ is $6.941\text{g/mol}$.

The formula to calculate moles of ${\text{Li}}_{\text{2}}\text{O}$ when $\text{Li}$ is limiting reagent is:

$\text{Moles}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\frac{\text{mass of}\text{Li}\left(\text{g}\right)}{\text{molecular mass}\text{of}\text{Li}\left(\text{g/mol}\right)}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{4\text{mol}\text{Li}}\right)\right]$                        (1)

Substitute $1.62\text{ g}$ for mass of $\text{Li}$ and $6.941\text{g/mol}$ for molecular mass of $\text{Li}$ in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\frac{1.62\text{ g}}{6.941\text{g/mol}}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{4\text{mol}\text{Li}}\right)\right]\\ =\left[\left(0.23339\text{mol}\right)\left(\frac{2\text{mol}{\text{Li}}_{\text{2}}\text{O}}{4\text{mol}\text{Li}}\right)\right]\\ =0.1166979\text{mol}\end{array}$

Conclusion

The moles of product formed when $1.62\text{ g}$ of lithium is mixed with $6.50\text{ g}$ of oxygen is $0.1166979\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

The mass of each reactant and product after the reaction is to be calculated.

Concept introduction:

The redox reaction can be classified into three types depending upon the number of reactants and products as follows:

1. Combination redox reaction

2. Decomposition redox reaction

3. Displacement redox reactions

Combination redox reactions are the reactions in which two or more reactants combine to form a single product. In displacement redox reactions, substances on both sides of the equation remain the same but the atoms exchange places in order to form the product while in decomposition reaction, one compound decompose to form one or more product.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

Answer to Problem 4.115P

The mass of $\text{Li}$, ${\text{O}}_2$ and ${\text{Li}}_{\text{2}}\text{O}$ after the reaction are 0, $4.63\text{g}$ and $3.49\text{g}$ respectively.

Explanation of Solution

Lithium is the limiting agent in the reaction. Hence, no moles of lithium will left after the completion of reaction.

The molecular mass of ${\text{Li}}_{\text{2}}\text{O}$ is $29.88\text{g/mol}$.

The formula to calculate mass of ${\text{Li}}_{\text{2}}\text{O}$ is:

$\text{Mass}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(\text{moles of}{\text{Li}}_{\text{2}}\text{O}\left(\text{mol}\right)\right)\left(\text{molecular}\text{mass}\text{of}{\text{Li}}_{\text{2}}\text{O}\left(\text{g/mol}\right)\right)\right]$          (3)

Substitute $0.1166979\text{mol}$ for moles of ${\text{Li}}_{\text{2}}\text{O}$ and $29.88\text{g/mol}$ for molecular mass of ${\text{Li}}_{\text{2}}\text{O}$ in the equation (3).

$\begin{array}{c}\text{Mass}\text{of}{\text{Li}}_{\text{2}}\text{O}=\left[\left(0.1166979\text{mol}\right)\left(29.88\text{g/mol}\right)\right]\\ =3.4869\text{g}\\ \approx 3.49\text{g}\end{array}$

The formula to calculate mass of ${\text{O}}_2$ reacted is:

$\text{Mass}\text{of}{\text{O}}_2\text{reacted}=\left[\begin{array}{l}\left(\frac{\text{mass of}\text{Li}\left(\text{g}\right)}{\text{molecular}\text{mass}\text{of}\text{Li}\left(\text{g/mol}\right)}\right)\left(\frac{1\text{mol}{\text{O}}_2}{4\text{mol}\text{Li}}\right)\\ \text{molecular}\text{mass}\text{of}{\text{O}}_2\left(\text{g/mol}\right)\end{array}\right]$                     (4)

Substitute $1.62\text{ g}$ for mass of $\text{Li}$, $6.941\text{g/mol}$ for molecular mass of $\text{Li}$ and $32.00\text{g/mol}$ for molecular mass of ${\text{O}}_2$ in the equation (4).

$\begin{array}{c}\text{Mass}\text{of}{\text{O}}_2\text{reacted}=\left[\left(\frac{1.62\text{ g}}{6.941\text{g/mol}}\right)\left(\frac{1\text{mol}{\text{O}}_2}{4\text{mol}\text{Li}}\right)\left(32.00\text{g/mol}\right)\right]\\ =1.867166\text{g}\end{array}$

The formula to calculate mass of remaining ${\text{O}}_2$ is:

${\text{Remaining O}}_2=\left(\text{Initial amount of}{\text{O}}_2\right)-\left(\text{Reacted amount of}{\text{O}}_2\right)$ (5)

Substitute $6.50\text{ g}$ for initial mass of ${\text{O}}_2$ and $1.867166\text{g}$ for reacted mass of ${\text{O}}_2$ in the equation (5).

$\begin{array}{c}{\text{Remaining O}}_2=\left(6.50\text{ g}\right)-\left(1.867166\text{g}\right)\\ =4.632834\text{g}\\ \approx 4.63\text{g}\end{array}$

Conclusion

The mass of $\text{Li}$, ${\text{O}}_2$ and ${\text{Li}}_{\text{2}}\text{O}$ after the reaction are 0, $4.63\text{g}$ and $3.49\text{g}$ respectively.