Chapter F, Problem H.6E

(a)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    ${\text{KClO}}_{\text{3}}(s)\stackrel{\Delta }{\to }\text{KCl}(s)+{\text{O}}_{\text{2}}(g)$

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.6E

Balanced chemical equation is ${\text{2KClO}}_{\text{3}}(s)\stackrel{\Delta }{\to }\text{2KCl}(s)+3{\text{O}}_{\text{2}}(g)$.

Explanation of Solution

Given skeletal equation is shown below;

    ${\text{KClO}}_{\text{3}}(s)\stackrel{\Delta }{\to }\text{KCl}(s)+{\text{O}}_{\text{2}}(g)$

Balancing oxygen atoms:  In the reactant side, there are three oxygen atoms while on the product side, there are two oxygen atoms.  Adding coefficient $2$ before ${\text{KClO}}_{\text{3}}$ and $3$ before ${\text{O}}_{\text{2}}$ balances the oxygen atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{2KClO}}_{\text{3}}(s)\stackrel{\Delta }{\to }\text{KCl}(s)+3{\text{O}}_{\text{2}}(g)$

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

    ${\text{2KClO}}_{\text{3}}(s)\stackrel{\Delta }{\to }\text{2KCl}(s)+3{\text{O}}_{\text{2}}(g)$

(b)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    ${\text{KClO}}_{\text{3}}(l)\stackrel{\Delta }{\to }\text{KCl}(s)+{\text{KClO}}_{\text{4}}(g)$

Concept Introduction:

Refer part (a).

Answer to Problem H.6E

Balanced chemical equation is ${\text{4KClO}}_{\text{3}}(l)\stackrel{\Delta }{\to }\text{KCl}(s)+3{\text{KClO}}_{\text{4}}(g)$.

Explanation of Solution

Given skeletal equation is shown below;

    ${\text{KClO}}_{\text{3}}(l)\stackrel{\Delta }{\to }\text{KCl}(s)+{\text{KClO}}_{\text{4}}(g)$

Balancing oxygen atoms:  In the reactant side, there are three oxygen atoms while on the product side, there are four oxygen atoms.  Adding coefficient $4$ before ${\text{KClO}}_{\text{3}}$ and $3$ before ${\text{KClO}}_{\text{4}}$ balances the oxygen atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation obtained is given as follows;

    ${\text{4KClO}}_{\text{3}}(l)\stackrel{\Delta }{\to }\text{KCl}(s)+3{\text{KClO}}_{\text{4}}(g)$

(c)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}(aq)+{\text{I}}_{\text{2}}(aq)\to \text{HI}(aq)+{\text{N}}_{\text{2}}(g)$

Concept Introduction:

Refer part (a).

Answer to Problem H.6E

Balanced chemical equation is ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}(aq)+2{\text{I}}_{\text{2}}(aq)\to 4\text{HI}(aq)+{\text{N}}_{\text{2}}(g)$.

Explanation of Solution

Given skeletal equation is shown below;

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}(aq)+{\text{I}}_{\text{2}}(aq)\to \text{HI}(aq)+{\text{N}}_{\text{2}}(g)$

Balancing hydrogen atoms:  In the reactant side, there are four hydrogen atoms while on the product side, there is only one hydrogen atom.  Adding coefficient $4$ before $\text{HI}$ balances the hydrogen atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}(aq)+{\text{I}}_{\text{2}}(aq)\to 4\text{HI}(aq)+{\text{N}}_{\text{2}}(g)$

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

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}(aq)+2{\text{I}}_{\text{2}}(aq)\to 4\text{HI}(aq)+{\text{N}}_{\text{2}}(g)$

(d)

Interpretation Introduction

Interpretation:

The given skeletal chemical equation has to be balanced.

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(l)$

Concept Introduction:

Refer part (a).

Answer to Problem H.6E

Balanced chemical equation is ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)\to 4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(l)$.

Explanation of Solution

Given skeletal equation is shown below;

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(l)$

Balancing phosphorus atoms:  In the reactant side, there are four phosphorus atoms while on the product side, there is only one phosphorus atom.  Adding coefficient $4$ before ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ balances the phosphorus atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)+{\text{H}}_{\text{2}}\text{O}(l)\to 4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(l)$

Balancing hydrogen atoms:  In the reactant side, there is only two hydrogen atoms while on the product side there are twelve hydrogen atoms.  Adding coefficient $6$ before ${\text{H}}_{\text{2}}\text{O}$ on the reactant side balances the hydrogen atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}(s)+6{\text{H}}_{\text{2}}\text{O}(l)\to 4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(l)$