Chapter 3, Problem 69E

(a)

Interpretation Introduction

Interpretation: The equation $\text{Cr}\left(s\right)+{\text{S}}_{\text{8}}\left(s\right)\to {\text{Cr}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)$ is to be balanced.

Concept introduction: Balancing any equation is important in order to know how many atoms of any element are participating in given reaction. Moreover, balancing of any equation is done to follow law of mass conservation and also for various stoichiometric calculations. Introducing stoichiometric coefficients for participants assist in balancing the reaction.

Answer to Problem 69E

The balanced equation is shown below.

  $\text{16Cr}\left(s\right)+3{\text{S}}_{\text{8}}\left(s\right)\to 8{\text{Cr}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)$

Explanation of Solution

The given equation is shown below.

  $\text{Cr}\left(s\right)+{\text{S}}_{\text{8}}\left(s\right)\to {\text{Cr}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)$

Balancing any equation involves equalizing the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{S}$ on the left-hand side is $8$ and that on right-hand side are $3$ . Therefore, in order to balance the $\text{S}$ atoms, add $3$ as the stoichiometric coefficient for ${\text{S}}_{\text{8}}\left(s\right)$ and $8$ as the stoichiometric coefficient of ${\text{Cr}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)$ . The equation obtained after doing this is as follows.

  $\text{Cr}\left(s\right)+3{\text{S}}_{\text{8}}\left(s\right)\to 8{\text{Cr}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)$ .

Similarly, the number of atoms of $\text{Cr}$ on the right-hand side is $16$ . Therefore, in order to balance the $\text{Cr}$ atoms, add $16$ as the stoichiometric coefficient for $\text{Cr}\left(s\right)$ on left-hand side. The equation obtained after doing this is as follows.

  $\text{16Cr}\left(\text{s}\right)+3{\text{S}}_{\text{8}}\left(\text{s}\right)\to 8{\text{Cr}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

The number of atoms of each element on each side is as follows.

  $\begin{array}{c}\text{Cr}=16\\ \text{S}=24\end{array}$

Since, the number of atoms of each element is equal on both sides; therefore the above equation is balanced.

(b)

Interpretation Introduction

Interpretation: The equation ${\text{NaHCO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is to be balanced.

Concept introduction: Balancing any equation is important in order to know how many atoms of any element are participating in given reaction. Moreover, balancing of any equation is done to follow law of mass conservation and also for various stoichiometric calculations. Introducing stoichiometric coefficients for participants assist in balancing the reaction.

Answer to Problem 69E

The balanced equation is shown below.

  ${\text{2NaHCO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The given equation is shown below.

  ${\text{NaHCO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{Na}$ on the right-hand side is $2$ and that on left-hand side is $1$ . Therefore, in order to balance the $\text{Na}$ atoms, add $2$ as the stoichiometric coefficient for ${\text{NaHCO}}_{\text{3}}\left(s\right)$ on left hand right. The equation obtained after doing this is as follows.

  ${\text{2NaHCO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The number of atoms of each element on each side is as follows.

  $\begin{array}{c}\text{Na}=\text{2}\\ \text{H}=2\\ \text{C}=2\\ \text{O}=6\end{array}$

Since, the number of atoms of each element is equal on both sides; therefore the above equation is balanced.

(c)

Interpretation Introduction

Interpretation: The equation ${\text{KClO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }\text{KCl}\left(s\right)+{\text{O}}_2\left(g\right)$ is to be balanced.

Concept introduction: Balancing any equation is important in order to know how many atoms of any element are participating in given reaction. Moreover, balancing of any equation is done to follow law of mass conservation and also for various stoichiometric calculations. Introducing stoichiometric coefficients for participants assist in balancing the reaction.

Answer to Problem 69E

The balanced equation is shown below.

  ${\text{2KClO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }\text{2KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

Explanation of Solution

The given equation is shown below.

  ${\text{KClO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }\text{KCl}\left(s\right)+{\text{O}}_2\left(g\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{O}$ on the right-hand side is $2$ and that on left-hand side is $3$ . Therefore, in order to balance the $\text{O}$ atoms, add $2$ as the stoichiometric coefficient for ${\text{KClO}}_{\text{3}}\left(s\right)$ on left-hand right and $3$ as the stoichiometric coefficient of ${\text{O}}_2\left(g\right)$ on right-hand side. The equation obtained after doing this is as follows.

  ${\text{2KClO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }\text{KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

In order to balance the number of atoms of $\text{K}$ and $\text{Cl}$ on both sides, add $2$ as the stoichiometric coefficient for $\text{KCl}\left(s\right)$ .The equation obtained after doing this is shown below.

  ${\text{2KClO}}_{\text{3}}\left(s\right)\stackrel{\text{Heat}}{\to }\text{2KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

The number of atoms of each element on each side is as follows.

  $\begin{array}{c}\text{K}=\text{2}\\ \text{Cl}=2\\ \text{O}=6\end{array}$

Since, the number of atoms of each element is equal on both sides; therefore the above equation is balanced.

(d)

Interpretation Introduction

Interpretation: The equation $\text{Eu}\left(s\right)+\text{HF}\left(g\right)\to {\text{EuF}}_3\left(s\right)+{\text{H}}_2\left(g\right)$ is to be balanced.

Concept introduction: Balancing any equation is important in order to know how many atoms of any element are participating in given reaction. Moreover, balancing of any equation is done to follow law of mass conservation and also for various stoichiometric calculations. Introducing stoichiometric coefficients for participants assist in balancing the reaction.

Answer to Problem 69E

The balanced equation is shown below.

  $\text{2Eu}\left(s\right)+6\text{HF}\left(g\right)\to 2{\text{EuF}}_3\left(s\right)+3{\text{H}}_2\left(g\right)$

Explanation of Solution

The given equation is shown below.

  $\text{Eu}\left(s\right)+\text{HF}\left(g\right)\to {\text{EuF}}_3\left(s\right)+{\text{H}}_2\left(g\right)$

In order to balance any equation, equalize the atoms of every element on reactant’s and product’s side.

The number of atoms of $\text{F}$ on the right-hand side is $3$ and that on left-hand side is $1$ . Therefore, in order to balance the $\text{F}$ atoms, add $3$ as the stoichiometric coefficient for $\text{HF}\left(g\right)$ on left hand right .The equation obtained after doing this is as follows.

  $\text{Eu}\left(s\right)+3\text{HF}\left(g\right)\to {\text{EuF}}_3\left(s\right)+{\text{H}}_2\left(g\right)$

The number of $\text{H}$ atoms on right-hand side is $2$ and that on left-hand side is $3$ . Therefore, in order to balance the number of atoms of $\text{H}$ atoms on both sides, add $3$ as the stoichiometric coefficient for ${\text{H}}_2\left(g\right)$ and change the stoichiometric coefficient of $\text{HF}\left(g\right)$ to $6$ . The equation obtained after doing this is as follows.

  $\text{Eu}\left(s\right)+6\text{HF}\left(g\right)\to {\text{EuF}}_3\left(s\right)+3{\text{H}}_2\left(g\right)$

Now again, the number of atoms of $\text{F}$ on the right-hand side is $3$ and that on left-hand side is $6$ . Therefore, in order to balance the $\text{F}$ atoms, add $2$ as the stoichiometric coefficient for ${\text{EuF}}_3\left(s\right)$ on right-hand right .The equation obtained after doing this is as follows.

  $\text{Eu}\left(s\right)+6\text{HF}\left(g\right)\to 2{\text{EuF}}_3\left(s\right)+3{\text{H}}_2\left(g\right)$

The number of atoms of $\text{Eu}$ on the right-hand side is $2$ and that on left-hand side is $1$ . Therefore, in order to balance the $\text{Eu}$ atoms, add $2$ as the stoichiometric coefficient for $\text{Eu}\left(s\right)$ on left-hand right .The equation obtained after doing this is shown below.

  $\text{2Eu}\left(s\right)+6\text{HF}\left(g\right)\to 2{\text{EuF}}_3\left(s\right)+3{\text{H}}_2\left(g\right)$

The number of atoms of each element on each side is as follows.

  $\begin{array}{c}\text{Eu}=\text{2}\\ \text{H}=6\\ \text{F}=6\end{array}$

Since, the number of atoms of each element is equal on both sides; therefore the above equation is balanced.