Chapter 3, Problem 3.34P

(a)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced.

    ${\text{F}}_{\text{2}}(g)+{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)\to {\text{AlF}}_{\text{3}}(s)+{\text{O}}_{\text{2}}(g)$

Concept Introduction:

Chemical reaction is represented in short form by using the molecular formulas of products and reactants in which an arrow is present between reactants and products is known as chemical equation.  Atoms are neither created nor destroyed in any chemical reaction.  Therefore, the resulting chemical equation written for the chemical reaction should also be balanced.

Steps followed to obtain balanced chemical equation:

• Chemical formulas of the products and reactants are written.  Individual reactants and products are separated using plus sign while the reactants are separated from products by an arrow.
• Elements that appear only once on the reactant side and product side is balanced first.  Subscript should not be changed in chemical formula while balancing coefficients can be added until the total number of atoms of an element becomes equal on reactant and product side.
• Other elements present also balanced in the same way by adding balancing coefficients.
• Final check has to be performed in order to check that each element that is present in the chemical equation is balanced.
• Symbols have to be added for solids, liquids gases and aqueous solutions to indicate the physical state of the known reactants and products.

Explanation of Solution

Chemical equation given is shown below.

    ${\text{F}}_{\text{2}}(g)+{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)\to {\text{AlF}}_{\text{3}}(s)+{\text{O}}_{\text{2}}(g)$

Above chemical equation is not balanced.  There are three oxygen atoms on the left side of the equation while there are two oxygen atoms on right side of equation.  Adding coefficient $2$ before ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ in the reactant side and $3$ before ${\text{O}}_{\text{2}}$ in the product side balances the oxygen atom.  Obtained equation after adding coefficient is given as shown below.

    ${\text{F}}_{\text{2}}(g)+2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)\to {\text{AlF}}_{\text{3}}(s)+3{\text{O}}_{\text{2}}(g)(notbalanced)$

In the above equation, there are four aluminium atoms in reactant side while there is only one aluminium atom on the product side.  Adding coefficient $4$ before ${\text{AlF}}_{\text{3}}$ in the product side balances the aluminium atom on both sides of the equation.  The chemical equation is given as shown below.

    ${\text{F}}_{\text{2}}(g)+2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)\to 4{\text{AlF}}_{\text{3}}(s)+3{\text{O}}_{\text{2}}(g)(notbalanced)$

In the above equation, there are two fluorine atoms in reactant side while there are twelve fluorine atoms on the product side.  Adding coefficient $6$ before ${\text{F}}_{\text{2}}$ in the reactant side balances the fluorine atom on both sides of the equation.  The balanced chemical equation is given as shown below.

    ${\text{6F}}_{\text{2}}(g)+2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)\to 4{\text{AlF}}_{\text{3}}(s)+3{\text{O}}_{\text{2}}(g)(balanced)$

(b)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced.

    ${\text{NH}}_{\text{3}}(g)+{\text{O}}_{\text{2}}(g)\to \text{NO}(g)+{\text{H}}_{\text{2}}\text{O}(l)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

    ${\text{NH}}_{\text{3}}(g)+{\text{O}}_{\text{2}}(g)\to \text{NO}(g)+{\text{H}}_{\text{2}}\text{O}(l)$

Above chemical equation is not balanced.  There are three hydrogen atoms on the left side of the equation while there are two hydrogen atoms on right side of equation.  Adding coefficient $2$ before ${\text{NH}}_{\text{3}}$ in the reactant side and $3$ before ${\text{H}}_{\text{2}}\text{O}$ in the product side balances the hydrogen atoms on both side of the equation.  The chemical equation is obtained as shown below.

    ${\text{2NH}}_{\text{3}}(g)+{\text{O}}_{\text{2}}(g)\to \text{NO}(g)+3{\text{H}}_{\text{2}}\text{O}(l)(notbalanced)$

In the above chemical equation, there are two nitrogen atoms on the reactant side while only one nitrogen atom is present on the product side.  Adding coefficient $2$ before $\text{NO}$ in the product side balances the nitrogen atoms on both sides of the equation.  The chemical equation obtained is shown as given below.

    ${\text{2NH}}_{\text{3}}(g)+{\text{O}}_{\text{2}}(g)\to 2\text{NO}(g)+3{\text{H}}_{\text{2}}\text{O}(l)(notbalanced)$

In the above chemical equation, there are two oxygen atoms on the reactant side while there are three oxygen atoms on the product side.  Adding coefficient $2.5$ before ${\text{O}}_{\text{2}}$ in the reactant side balances the oxygen atoms on both sides of the equation.  The balanced chemical equation is obtained as shown below.

    $\begin{array}{l}{\text{2NH}}_{\text{3}}(g)+2.5{\text{O}}_{\text{2}}(g)\to 2\text{NO}(g)+3{\text{H}}_{\text{2}}\text{O}(l)\\ {\text{4NH}}_{\text{3}}(g)+5{\text{O}}_{\text{2}}(g)\to 4\text{NO}(g)+6{\text{H}}_{\text{2}}\text{O}(l)(balanced)\end{array}$

(c)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}(l)+{\text{O}}_{\text{2}}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(l)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}(l)+{\text{O}}_{\text{2}}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(l)$

Above chemical equation is not balanced.  There are six carbon atoms on the reactant side while only one carbon atom is present on the product side.  Adding coefficient $6$ before ${\text{CO}}_{\text{2}}$ in the product side balances the carbon atoms on both sides of the equation.  The chemical equation obtained is shown as given below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}(l)+{\text{O}}_{\text{2}}(g)\to 6{\text{CO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(l)(notbalanced)$

In the above chemical equation, there are six hydrogen atoms on the reactant side while only two hydrogen atoms are present in the product side.  Adding coefficient $3$ before ${\text{H}}_{\text{2}}\text{O}$ in the product side balances the hydrogen atoms on both sides of the equation.  The chemical equation obtained is shown as given below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}(l)+{\text{O}}_{\text{2}}(g)\to 6{\text{CO}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}\text{O}(l)(notbalanced)$

In the above chemical equation, there are two oxygen atoms on the reactant side while there are fifteen oxygen atoms on the product side.  Adding coefficient $7.5$ before ${\text{O}}_{\text{2}}$ in the reactant side balances the oxygen atoms on both sides of the equation.  The balanced chemical equation obtained is shown as given below.

    $\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}(l)+7.5{\text{O}}_{\text{2}}(g)\to 6{\text{CO}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}\text{O}(l)\\ {\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}(l)+15{\text{O}}_{\text{2}}(g)\to 12{\text{CO}}_{\text{2}}(g)+6{\text{H}}_{\text{2}}\text{O}(l)(balanced)\end{array}$

(d)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced.

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

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

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

Above chemical equation is not balanced.  There is one sulfur atom present on the reactant side while three sulfur atoms are present on the product side.  Adding coefficient $3$ before ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ balances sulfur atoms on both sides of the equation.  The obtained equation is given as shown below.

    ${\text{3H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{Al}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3}}(s)\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+{\text{CO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(l)(notbalanced)$

In the above equation, there are six hydrogen atoms on the reactant side while there are two hydrogen atoms on the product side.  Adding coefficient $3$ before ${\text{H}}_{\text{2}}\text{O}$ balances the hydrogen atoms on both sides of the equation.  The chemical equation obtained is given as shown below.

    ${\text{3H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{Al}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3}}(s)\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+{\text{CO}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}\text{O}(l)(notbalanced)$

In the above equation, there are three carbon atoms on the reactant side while there is only one carbon atom on the product side.  Adding coefficient $3$ before ${\text{CO}}_{\text{2}}$ balances the carbon atoms on both sides of the equation.  The balanced chemical equation obtained is given as shown below.

    ${\text{3H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{Al}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3}}(s)\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(aq)+3{\text{CO}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}\text{O}(l)(balanced)$