Chapter 3, Problem 45QRT

(a)

Interpretation Introduction

Interpretation:

The product and the type of the reaction for $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{ + 3HNO}}_{\text{3}}$ should be identified. Also the reaction has to be balanced.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the ions combine in the solution is called precipitation reaction.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

Gas forming reaction: The reaction of acid and metal carbonates which produce carbonic acid. The carbonic acid decomposes which gives water and carbon dioxide.

Oxidation - reduction reaction: The electrons are transferred to one to other is called oxidation reduction reaction.

Balanced reaction:

Balanced reaction is a chemical reaction in which number of atoms for each element in the reaction and the total charge are same on both reactant side and the product side.

Steps in balancing the information

• Step 1: Write the unbalanced equation
• Step 2: Find the coefficient to balance the equation.
• The coefficient should be reduced to the smallest whole number.

Explanation of Solution

Given,

  $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{ + 3HNO}}_{\text{3}}$

The product of the reaction is shown below,

  $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{ + HNO}}_{\text{3}}\left(\text{aq}\right)\to \text{ Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{+ H}}_{\text{2}}\text{O}$

The given reaction is Acid - base reaction. In this reaction Iron(III) oxide-hydroxide reaction with nitric acid gives Iron(III) nitrate and water. Formation of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is salt from the iron cation from the base of Iron(III) oxide-hydroxide and nitrate anion from the nitric acid and formation of water is also the product. Therefore, it is Acid - base reaction.

  $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{ + HNO}}_{\text{3}}\left(\text{aq}\right)\to \text{ Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{+ H}}_{\text{2}}\text{O}$

Balance the equation,

When balancing the equation, we should not alter the subscripts and we can change coefficients. There are three nitrogen atoms in the right side and one nitrogen atoms in the left side. Therefore, three molecules of nitric acid is added to left side of reaction. Therefore, the balanced equation is given below.

  $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{ + 3HNO}}_{\text{3}}\left(\text{aq}\right)\to \text{ Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{+ H}}_{\text{2}}\text{O}$

Now, there are six hydrogen atoms in the left side and two hydrogen atoms in the right side. Therefore, three molecules of water are added to right side of reaction. Therefore, the balanced equation is given below.

  $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{ + 3HNO}}_{\text{3}}\left(\text{aq}\right)\to \text{ Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{+ 3H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

The product and the type of the reaction for ${\text{FeCO}}_{\text{3}}\left(\text{s}\right){\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)$ should be identified.  Also the reaction has to be balanced.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

  ${\text{FeCO}}_{\text{3}}\left(\text{s}\right){\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)$

The product of the reaction is shown below,

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

The given reaction is gas forming reaction, the given compound is Iron(II) carbonate and sulfuric acid. In this reaction Iron(II) carbonate reaction with sulfuric acid gives Iron(II) sulfate, water and carbon dioxide.

The reaction of sulfuric acid and Iron(II) carbonate which produce carbonic acid. The carbonic acid decomposes which gives water and carbon dioxide. Therefore, the reaction gas forming reaction.

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

Balance the equation,

When balancing the equation, we should not alter the subscripts and we can change coefficients. The left side of the molecule is equal to right side of the molecules.

Therefore, the balanced equation is given below.

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

(c)

Interpretation Introduction

Interpretation:

The product and the type of the reaction for ${\text{FeCl}}_{\text{2}}\left(\text{aq}\right)\text{+}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{aq}\right)$ should be identified.  Also the reaction has to be balanced.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

  ${\text{FeCl}}_{\text{2}}\left(\text{aq}\right)\text{+}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{aq}\right)$

The product of the reaction is shown below,

  ${\text{FeCl}}_{\text{2}}\left(\text{aq}\right)\text{+}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{aq}\right)\to \text{ FeS}\left(\text{S}\right){\text{ + NH}}_{\text{4}}\text{Cl}\left(\text{aq}\right)$

The given reaction is precipitation reaction, the given compound is Iron(II) chloride and Ammonium sulfide which is soluble in water. In this reaction Iron(II) chloride reaction with Ammonium sulfide gives Iron(II) sulfide and ammonium chloride.

  ${\text{FeCl}}_{\text{2}}\left(\text{aq}\right)\text{+}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{aq}\right)\to \text{ FeS}\left(\text{S}\right){\text{ + NH}}_{\text{4}}\text{Cl}\left(\text{aq}\right)$

Balance the equation,

when balancing the equation, we should not alter the subscripts and we can change coefficients. The left side of the molecule is equal to right side of the molecules.

Therefore, the balanced equation is given below.

  ${\text{FeCl}}_{\text{2}}\left(\text{aq}\right)\text{+}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{aq}\right)\to \text{ FeS}\left(\text{S}\right){\text{ + NH}}_{\text{4}}\text{Cl}\left(\text{aq}\right)$

(d)

Interpretation Introduction

Interpretation:

The product and the type of the reaction for $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)$ should be identified.  Also the reaction has to be balanced.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)$

The product of the reaction is shown below,

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{+ Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to {\text{FeCO}}_{\text{3}}\left(\text{s}\right){\text{ + NaNO}}_{\text{3}}\left(\text{aq}\right)$

The given reaction is precipitation reaction, the given compound is Iron(III) nitrate and sodium carbonate which is soluble in water. In this reaction Iron(III) nitrate reaction with sodium carbonate gives Iron(II) carbonate and sodium nitrate.

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{+ Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to {\text{FeCO}}_{\text{3}}\left(\text{s}\right){\text{ + NaNO}}_{\text{3}}\left(\text{aq}\right)$

Balance the equation,

when balancing the equation, we should not alter the subscripts and we can change coefficients. There are two nitrogen atoms in the left side and one nitrogen atoms in the right side. Therefore, two molecules of sodium nitrate is added to right side of reaction.

The left side of the molecule is equal to right side of the molecules, therefore, the balanced equation is given below.

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to {\text{FeCO}}_{\text{3}}\left(\text{s}\right){\text{ + 2 NaNO}}_{\text{3}}\left(\text{aq}\right)$