Chapter 7, Problem 40E

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the decomposition reaction of solid chromium $\left(\text{III}\right)$ carbonate is to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atoms or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Answer to Problem 40E

The balanced equation for the decomposition reaction of solid chromium $\left(\text{III}\right)$ carbonate is shown below.

${\text{Cr}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\left(s\right)\to {\text{Cr}}_2{\text{O}}_{\text{3}}\left(s\right)+3{\text{CO}}_{\text{2}}\left(g\right)$

Explanation of Solution

Solid chromium $\left(\text{III}\right)$ carbonate undergoes a decomposition reaction to produce solid chromium $\left(\text{III}\right)$ oxide and carbon dioxide gas.

The unbalanced equation for the decomposition reaction of solid chromium $\left(\text{III}\right)$ carbonate is shown below.

${\text{Cr}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\left(s\right)\to {\text{Cr}}_2{\text{O}}_{\text{3}}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

The numbers of atoms of elements on the left side of the reaction are not equal to that on the right side of the reaction. So, the reaction is unbalanced.

A coefficient of $3$ is placed in front of ${\text{CO}}_{\text{2}}$ to balance the chemical equation.

Therefore, the balanced equation for the decomposition reaction of solid chromium $\left(\text{III}\right)$ carbonate is shown below.

${\text{Cr}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\left(s\right)\to {\text{Cr}}_2{\text{O}}_{\text{3}}\left(s\right)+3{\text{CO}}_{\text{2}}\left(g\right)$

The numbers of atoms of each element on the left side of the reaction are equal to that on the right side of the reaction. Therefore, the reaction is balanced.

Conclusion

The balanced equation for the decomposition reaction of solid chromium $\left(\text{III}\right)$ carbonate is shown below.

${\text{Cr}}_2{\left({\text{CO}}_{\text{3}}\right)}_3\left(s\right)\to {\text{Cr}}_2{\text{O}}_{\text{3}}\left(s\right)+3{\text{CO}}_{\text{2}}\left(g\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the decomposition reaction of solid lead $\left(\text{IV}\right)$ carbonate is to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atoms or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Answer to Problem 40E

The balanced equation for the decomposition reaction of solid lead $\left(\text{IV}\right)$ carbonate is shown below.

$\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_2\left(s\right)\to {\text{PbO}}_{\text{2}}\left(s\right)+2{\text{CO}}_{\text{2}}\left(g\right)$

Explanation of Solution

Solid lead $\left(\text{IV}\right)$ carbonate undergoes a decomposition reaction to solid lead $\left(\text{IV}\right)$ oxide and carbon dioxide gas.

The unbalanced equation for the decomposition reaction of solid lead $\left(\text{IV}\right)$ carbonate is shown below.

$\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_2\left(s\right)\to {\text{PbO}}_{\text{2}}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

The numbers of atoms of elements on the left side of the reaction are not equal to the right side of the reaction. Therefore, the reaction is unbalanced.

A coefficient of $2$ is placed in front of ${\text{CO}}_{\text{2}}$ to balance the chemical equation.

Therefore, the balanced equation for the decomposition reaction of solid lead $\left(\text{IV}\right)$ carbonate is shown below.

$\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_2\left(s\right)\to {\text{PbO}}_{\text{2}}\left(s\right)+2{\text{CO}}_{\text{2}}\left(g\right)$

The numbers of atoms of each element on the left side of the reaction are equal to that on the right side of the reaction. Therefore, the reaction is balanced.

Conclusion

The balanced equation for the decomposition reaction of solid lead $\left(\text{IV}\right)$ carbonate is shown below.

$\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_2\left(s\right)\to {\text{PbO}}_{\text{2}}\left(s\right)+2{\text{CO}}_{\text{2}}\left(g\right)$