Chapter 7, Problem 78E

Interpretation Introduction

(a)

Interpretation:

The reaction ${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is to be completed and balanced.

Concept introduction:

The process in which the reactants combine to form a new product is called a chemical reaction. The chemical and physical properties of the products are different from those of the reactants. The different types of chemical reactions are combination reaction, decomposition reaction, and displacement reaction.

Answer to Problem 78E

The reaction ${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is completed and balanced as shown below. ${\text{H}}_2{\text{SO}}_4\left(aq\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{4}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The reaction of an acid with a base results in the formation of salt and water. This reaction is known as the neutralization reaction.

Therefore, the chemical equation for the reaction ${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is given below.

${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{4}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of atoms on either side of the equation is not the same. Therefore, the coefficient $2$ is added before $\text{LiOH}\left(aq\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ to balance the equation. Therefore, the balanced equation is given below.

${\text{H}}_2{\text{SO}}_4\left(aq\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{4}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The reaction ${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is completed and balanced as shown below. ${\text{H}}_2{\text{SO}}_4\left(aq\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{4}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The reaction ${\text{H}}_2{\text{SO}}_3\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is to be completed and balanced.

Concept introduction:

The process in which the reactants combine to form a new product is called a chemical reaction. The chemical and physical properties of the products are different from those of the reactants. The different types of chemical reactions are combination reaction, decomposition reaction, and displacement reaction.

Answer to Problem 78E

The reaction ${\text{H}}_2{\text{SO}}_3\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is completed and balanced as shown below. ${\text{H}}_2{\text{SO}}_3\left(aq\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{3}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The reaction of an acid with a base results in the formation of salt and water. This reaction is known as the neutralization reaction.

Therefore, the chemical equation for the reaction ${\text{H}}_2{\text{SO}}_3\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is given below.

${\text{H}}_2{\text{SO}}_3\left(aq\right)+\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The numbers of atoms on either side of the equation are not the same. Therefore, the coefficient $2$ is added before $\text{LiOH}\left(aq\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ to balance the equation. Therefore, the balanced equation is given below.

${\text{H}}_2{\text{SO}}_3\left(aq\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{3}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The reaction ${\text{H}}_2{\text{SO}}_3\left(aq\right)+\text{LiOH}\left(aq\right)\to$ is completed and balanced as shown below. ${\text{H}}_2{\text{SO}}_3\left(aq\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_2{\text{SO}}_{\text{3}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$