Chapter 9, Problem 71A

Interpretation Introduction

(a)

Interpretation:

The skeleton equation of the reaction needs to be written.

$\text{iron}\left(\text{s}\right)\text{+}\text{fluorine}\left(\text{g}\right)\to \text{iron}\left(\text{III}\right)\text{fluoride}\left(\text{s}\right)$

Concept introduction:

In a chemical reaction, the reactants are written on the left side of the equation and the product is written on the right side. The product and the reactants contain an arrow between them showing the proceeding of the reaction from the reactants to the products. There are many ways to show a chemical reaction, skeleton equation, word equation and chemical equation.

Answer to Problem 71A

The skeleton equation of the reaction,

$\text{iron}\left(\text{s}\right)\text{+}\text{fluorine}\left(\text{g}\right)\to \text{iron}\left(\text{III}\right)\text{fluoride}\left(\text{s}\right)$ is,

$\text{Fe}\left(s\right){\text{+F}}_{\text{2}}\left(g\right)\to {\text{FeF}}_{\text{3}}\left(s\right)$

Explanation of Solution

To determine the skeleton equation for a reaction, one must know the chemical symbols of elements and compounds. A skeleton equation shows the reactants and products using the chemical formula of them and the balancing of the reaction is not required.

The formula of iron and fluorine is $\text{Fe}$ and ${\text{F}}_{\text{2}}$ respectively. So, the skeleton equation is,

$\text{Fe}\left(s\right){\text{+F}}_{\text{2}}\left(g\right)\to {\text{FeF}}_{\text{3}}\left(s\right)$

(b)

Interpretation Introduction

Interpretation:

The skeleton equation of the reaction needs to be written.

$\text{sulfur}\text{trioxide}\left(\text{g}\right)\text{+}\text{water}\left(\text{l}\right)\to \text{sulfuric}\text{acid}\left(\text{aq}\right)$

Concept introduction:

In a chemical reaction, the reactants are written on the left side of the equation and the product is written on the right side. The product and the reactants contain an arrow between them showing the proceeding of the reaction from the reactants to the products. There are many ways to show a chemical reaction, skeleton equation, word equation and chemical equation.

Answer to Problem 71A

The skeleton equation of the reaction,

$\text{sulfur}\text{trioxide}\left(\text{g}\right)\text{+}\text{water}\left(\text{l}\right)\to \text{sulfuric}\text{acid}\left(\text{aq}\right)$ is,

${\text{SO}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)$

Explanation of Solution

To determine the skeleton equation for a reaction, one must know the chemical symbols of elements and compounds. A skeleton equation shows the reactants and products using the chemical formula of them and the balancing of the reaction is not required.

The formula of Sulphur, oxygen and water is S, ${\text{O}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ respectively. So, the skeleton equation is,

${\text{SO}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)$

(c)

Interpretation Introduction

Interpretation:

The skeleton equation of the reaction needs to be written.

$\text{sodium}\left(\text{s}\right)\text{+}\text{magnesium}\text{iodide}\left(\text{aq}\right)\to \text{sodium}\text{iodide}\left(\text{aq}\right)\text{+}\text{magnesium}\left(\text{s}\right)$

Concept introduction:

In a chemical reaction, the reactants are written on the left side of the equation and the product is written on the right side. The product and the reactants contain an arrow between them showing the proceeding of the reaction from the reactants to the products. There are many ways to show a chemical reaction, skeleton equation, word equation and chemical equation.

Answer to Problem 71A

The skeleton equation of the reaction,

$\text{sodium}\left(\text{s}\right)\text{+}\text{magnesium}\text{iodide}\left(\text{aq}\right)\to \text{sodium}\text{iodide}\left(\text{aq}\right)\text{+}\text{magnesium}\left(\text{s}\right)$ is,

$\text{Na}\left(\text{s}\right)\text{+MgI}\left(\text{aq}\right)\to \text{NaI}\left(\text{aq}\right)\text{+Mg}\left(\text{s}\right)$

Explanation of Solution

To determine the skeleton equation for a reaction, one must know the chemical symbols of elements and compounds. A skeleton equation shows the reactants and products using the chemical formula of them and the balancing of the reaction is not required.

The formula of sodium, magnesium and iodine is Na, Mg and I respectively.

So, the skeleton equation is,

$\text{Na}\left(\text{s}\right)\text{+MgI}\left(\text{aq}\right)\to \text{NaI}\left(\text{aq}\right)\text{+Mg}\left(\text{s}\right)$

(d)

Interpretation Introduction

Interpretation:

The skeleton equation of the reaction needs to be written.

$\text{vanadium}\left(\text{s}\right)\text{+}\text{oxygen}\left(\text{g}\right)\to \text{vanadium}\left(\text{v}\right)\text{oxide}\left(\text{s}\right)$

Concept introduction:

In a chemical reaction, the reactants are written on the left side of the equation and the product is written on the right side. The product and the reactants contain an arrow between them showing the proceeding of the reaction from the reactants to the products. There are many ways to show a chemical reaction, skeleton equation, word equation and chemical equation.

Answer to Problem 71A

The skeleton equation of the reaction,

$\text{vanadium}\left(\text{s}\right)\text{+}\text{oxygen}\left(\text{g}\right)\to \text{vanadium}\left(\text{v}\right)\text{oxide}\left(\text{s}\right)$ is,

$\text{V}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}\text{(excess)}\stackrel{\Delta }{\to }{\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\left(\text{s}\right)$

Explanation of Solution

To determine the skeleton equation for a reaction, one must know the chemical symbols of elements and compounds. A skeleton equation shows the reactants and products using the chemical formula of them and the balancing of the reaction is not required.

The formula of vanadium, oxygen and vanadium oxide is, $\text{V}$, ${\text{O}}_{\text{2}}$ and ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}$ respectively.

So, the skeleton equation is,

$\text{V}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}\text{(excess)}\stackrel{\Delta }{\to }{\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\left(\text{s}\right)$