Chapter F, Problem H.8E

(a)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the reaction of molecular oxygen with ${\text{CuFeS}}_{\text{2}}$ to produce copper(II) sulfide, iron(II) oxide, and sulfur dioxide gas.

Concept Introduction:

Chemical equation is a short form representation of a chemical reaction with all the required conditions necessary for the reaction to take place.  As the atoms are neither created nor destroyed in a chemical reaction, the number of atoms of same element has to be equal on both sides of the equation.  This is known as balanced chemical equation.  Coefficients can be changed but not the subscript.

Answer to Problem H.8E

Balanced chemical equation is ${\text{2CuFeS}}_{\text{2}}(s)+3{\text{O}}_{\text{2}}(g)\to 2\text{CuS}(s)+2\text{FeO}(s)+2{\text{SO}}_{\text{2}}(g)$.

Explanation of Solution

The reaction given is molecular oxygen with ${\text{CuFeS}}_{\text{2}}$ to produce copper(II) sulfide, iron(II) oxide, and sulfur dioxide gas.  The skeletal equation for the reaction can be represented as shown below;

    ${\text{CuFeS}}_{\text{2}}(s)+{\text{O}}_{\text{2}}(g)\to \text{CuS}(s)+\text{FeO}(s)+{\text{SO}}_{\text{2}}(g)$

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms while on the product side, there are three oxygen atoms.  Adding coefficient $1.5$ before ${\text{O}}_{\text{2}}$ balances the oxygen atoms on both sides of the equation.  This step balances the other atoms also.  Therefore, the balanced chemical equation can be written as follows;

    ${\text{CuFeS}}_{\text{2}}(s)+1.5{\text{O}}_{\text{2}}(g)\to \text{CuS}(s)+\text{FeO}(s)+{\text{SO}}_{\text{2}}(g)$

In order to obtain in whole numbers, the equation is multiplied by $2$.  Thus, the balanced chemical equation can be given as follows;

    ${\text{2CuFeS}}_{\text{2}}(s)+3{\text{O}}_{\text{2}}(g)\to 2\text{CuS}(s)+2\text{FeO}(s)+2{\text{SO}}_{\text{2}}(g)$

(b)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the reaction of silicon dioxide with carbon to form silicon carbide and carbon monoxide gas.

Concept Introduction:

Refer part (a).

Answer to Problem H.8E

Balanced chemical equation is ${\text{SiO}}_{\text{2}}(s)+3\text{C}(s)\stackrel{\text{2000°C}}{\to }\text{SiC}(s)+\text{2CO}(g)$.

Explanation of Solution

The reaction given is silicon dioxide with carbon to form silicon carbide and carbon monoxide gas..  The skeletal equation for the reaction can be represented as shown below;

    ${\text{SiO}}_{\text{2}}(s)+\text{C}(s)\stackrel{\text{2000°C}}{\to }\text{SiC}(s)+CO(g)$

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms while on the product side, there is only one oxygen atom.  Adding coefficient $2$ before $\text{CO}$ balances the oxygen atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    ${\text{SiO}}_{\text{2}}(s)+\text{C}(s)\stackrel{\text{2000°C}}{\to }\text{SiC}(s)+\text{2CO}(g)$

Balancing carbon atoms:  In the reactant side, there is one carbon atom while on the product side, there are three carbon atoms.  Adding coefficient $3$ before $\text{C}$ balances the carbon atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    ${\text{SiO}}_{\text{2}}(s)+3\text{C}(s)\stackrel{\text{2000°C}}{\to }\text{SiC}(s)+\text{2CO}(g)$

(c)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the reaction of hydrogen and nitrogen to produce ammonia.

Concept Introduction:

Refer part (a).

Answer to Problem H.8E

Balanced chemical equation is ${\text{3H}}_{\text{2}}(g)+{\text{N}}_{\text{2}}(g)\to 2{\text{NH}}_{\text{3}}(g)$.

Explanation of Solution

The reaction given is hydrogen and nitrogen to produce ammonia.  The skeletal equation for the reaction can be represented as shown below;

    ${\text{H}}_{\text{2}}(g)+{\text{N}}_{\text{2}}(g)\to {\text{NH}}_{\text{3}}(g)$

Balancing nitrogen atoms:  In the reactant side, there are two nitrogen atoms while on the product side, there is only one nitrogen atom.  Adding coefficient $2$ before ${\text{NH}}_{\text{3}}$ balances the nitrogen atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    ${\text{H}}_{\text{2}}(g)+{\text{N}}_{\text{2}}(g)\to 2{\text{NH}}_{\text{3}}(g)$

Balancing hydrogen atoms:  In the reactant side, there are two hydrogen atoms while on the product side, there are six hydrogen atoms.  Adding coefficient $3$ before ${\text{H}}_{\text{2}}$ balances the hydrogen atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    ${\text{3H}}_{\text{2}}(g)+{\text{N}}_{\text{2}}(g)\to 2{\text{NH}}_{\text{3}}(g)$

(d)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the reaction of oxygen with aqueous hydrobromic acid to form liquid water and bromine.

Concept Introduction:

Refer part (a).

Answer to Problem H.8E

Balanced chemical equation is $\text{4HBr}(aq)+{\text{O}}_{\text{2}}(g)\to 2{\text{H}}_{\text{2}}\text{O}(l)+2{\text{Br}}_{\text{2}}(l)$.

Explanation of Solution

The reaction given is oxygen with aqueous hydrobromic acid to form liquid water and bromine.  The skeletal equation for the reaction can be represented as shown below;

    $\text{HBr}(aq)+{\text{O}}_{\text{2}}(g)\to {\text{H}}_{\text{2}}\text{O}(l)+{\text{Br}}_{\text{2}}(l)$

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms while on the product side, there is one oxygen atom.  Adding coefficient $2$ before ${\text{H}}_{\text{2}}\text{O}$ balances the oxygen atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    $\text{HBr}(aq)+{\text{O}}_{\text{2}}(g)\to 2{\text{H}}_{\text{2}}\text{O}(l)+{\text{Br}}_{\text{2}}(l)$

Balancing hydrogen atoms:  In the reactant side, there is one hydrogen atom while on the product side, there are four hydrogen atoms.  Adding coefficient $4$ before $\text{HBr}$ balances the hydrogen atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    $\text{4HBr}(aq)+{\text{O}}_{\text{2}}(g)\to 2{\text{H}}_{\text{2}}\text{O}(l)+{\text{Br}}_{\text{2}}(l)$

Balancing bromine atoms:  In the reactant side, there are four bromine atoms while on the product side, there are two bromine atoms.  Adding coefficient $2$ before ${\text{Br}}_{\text{2}}$ balances the bromine atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    $\text{4HBr}(aq)+{\text{O}}_{\text{2}}(g)\to 2{\text{H}}_{\text{2}}\text{O}(l)+2{\text{Br}}_{\text{2}}(l)$