Chapter 3, Problem 3.81P

(a)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement when hydrogen sulfide forms sulfur dioxide and water vapor on reaction with oxygen should be written.

Concept introduction: In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.81P

The balanced equation is written as follows:

  ${\text{2H}}_2\text{S}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2{\text{SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The formula ofgallium oxide is ${\text{H}}_2\text{S}$ .The unbalanced equation that needs to be balanced is as follows:

  ${\text{H}}_2\text{S}\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Since, left side has 2 $\text{O}$ while right side has 3 $\text{O}$ so first the stoichiometric coefficient of ${\text{O}}_2$ on the left is made 3/2.

  ${\text{H}}_2\text{S}\left(s\right)+\frac{3}{2}{\text{O}}_2\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Multiple the entire equation by 2.

  ${\text{2H}}_2\text{S}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2{\text{SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Since above equation has equal atoms on right and left side so the equation is now balanced.

(b)

Interpretation Introduction

Interpretation: The balanced equation corresponding to statement when potassium chlorate is heated to form two potassium chloride and potassium perchlorate should be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.81P

The balanced equation is written as follows:

  ${\text{4KClO}}_{\text{3}}\left(s\right)\stackrel{\Delta }{\to }3{\text{KClO}}_4\left(s\right)+\text{KCl}\left(s\right)$

Explanation of Solution

The formula ofpotassium chlorate and potassium perchlorate are ${\text{KClO}}_{\text{3}}$ and ${\text{KClO}}_4$ . The unbalanced equation that needs to be balanced is as follows:

  ${\text{KClO}}_{\text{3}}\left(s\right)\stackrel{\Delta }{\to }{\text{KClO}}_4\left(s\right)+\text{KCl}\left(s\right)$

Since, left side has 3 $\text{O}$ while right side has 4 $\text{O}$ so place 4 as the coefficient of ${\text{KClO}}_{\text{3}}$ and 3 as coefficient of ${\text{KClO}}_4$ so as to make equal $\text{O}$ on both side.

  ${\text{4KClO}}_{\text{3}}\left(s\right)\stackrel{\Delta }{\to }3{\text{KClO}}_4\left(s\right)+\text{KCl}\left(s\right)$

Since above equation has equal atoms on each side so the equation is now balanced.

(c)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement that when hydrogen gas is passed on $\text{iron}\left(\text{III}\right)$ oxide to form iron metal and water vapor formshould be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.81P

The balanced equation is written as follows:

  ${\text{3H}}_{\text{2}}\left(g\right)+{\text{Fe}}_{\text{3}}{\text{O}}_3\left(s\right)\to 3{\text{H}}_{\text{2}}\text{O}\left(g\right)+2\text{Fe}\left(s\right)$

Explanation of Solution

The formula ofcalcium chloride, sodium phosphate are ${\text{CaCl}}_{\text{2}}$ and ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ respectively while the formula ofcalcium phosphate and sodium chloride are ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and $\text{NaCl}$ respectively.

The unbalanced equation that needs to be balanced is as follows:

  ${\text{H}}_{\text{2}}\left(g\right)+{\text{Fe}}_{\text{3}}{\text{O}}_3\left(s\right)\to {\text{H}}_{\text{2}}\text{O}\left(g\right)+\text{Fe}\left(s\right)$

Since, left side has 3 $\text{O}$ while right side has 4 $\text{O}$ so place 4 as the coefficient of ${\text{KClO}}_{\text{3}}$ and 3 as coefficient of ${\text{KClO}}_4$ so as to make equal $\text{O}$ on both side.

  ${\text{H}}_{\text{2}}\left(g\right)+{\text{Fe}}_{\text{3}}{\text{O}}_3\left(s\right)\to 3{\text{H}}_{\text{2}}\text{O}\left(g\right)+\text{Fe}\left(s\right)$

Place 3 as the coefficient of ${\text{H}}_{\text{2}}$ so as to make equal 6 $\text{H}$ on each side.

  ${\text{3H}}_{\text{2}}\left(g\right)+{\text{Fe}}_{\text{3}}{\text{O}}_3\left(s\right)\to 3{\text{H}}_{\text{2}}\text{O}\left(g\right)+\text{Fe}\left(s\right)$

Place 2 as the coefficient of $\text{Fe}$ so as to make 2 $\text{Fe}$ on each side.

  ${\text{3H}}_{\text{2}}\left(g\right)+{\text{Fe}}_{\text{3}}{\text{O}}_3\left(s\right)\to 3{\text{H}}_{\text{2}}\text{O}\left(g\right)+2\text{Fe}\left(s\right)$

Since above equation has equal atoms on right and left side so the equation is now balanced.

(d)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement that when ethane undergoes combustion should be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.81P

The balanced equation is written as follows:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+7{\text{O}}_2\left(g\right)\to 6{\text{H}}_{\text{2}}\text{O}\left(g\right)+4{\text{CO}}_{\text{2}}\left(g\right)$

Explanation of Solution

The unbalanced equation that needs to be balanced is as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Place 3 as coefficient of ${\text{H}}_{\text{2}}\text{O}$ and 2 as coefficient of ${\text{CO}}_{\text{2}}$ so as to make equal $\text{H}$ and $\text{C}$ on each side.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+{\text{O}}_2\left(g\right)\to 3{\text{H}}_{\text{2}}\text{O}\left(g\right)+2{\text{CO}}_{\text{2}}\left(g\right)$

Since, left side has 2 $\text{O}$ while right side has 7 $\text{O}$ so place 7/2 as coefficient of ${\text{O}}_2$ to make equal $\text{O}$ on both side.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+\frac{7}{2}{\text{O}}_2\left(g\right)\to 3{\text{H}}_{\text{2}}\text{O}\left(g\right)+2{\text{CO}}_{\text{2}}\left(g\right)$

Multiple the entire equation by 2.

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+7{\text{O}}_2\left(g\right)\to 6{\text{H}}_{\text{2}}\text{O}\left(g\right)+4{\text{CO}}_{\text{2}}\left(g\right)$

Since above equation has equal atoms on right and left side so the equation is now balanced.

(e)

Interpretation Introduction

Interpretation: The balanced equation corresponding to the statement hen $\text{iron}\left(\text{II}\right)$ chloride is converted to $\text{iron}\left(\text{III}\right)$ ) fluoride should be written.

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.81P

The balanced equation is written as follows:

  ${\text{2FeCl}}_2\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_3\left(s\right)+3{\text{Cl}}_{\text{2}}\left(g\right)$

Explanation of Solution

The unbalanced equation when $\text{iron}\left(\text{II}\right)$ chloride is converted to $\text{iron}\left(\text{III}\right)$ ) fluoride is as follows:

  ${\text{FeCl}}_2\left(s\right)+{\text{ClF}}_{\text{3}}\left(g\right)\to {\text{FeF}}_3\left(s\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

Place 3 as coefficient of ${\text{ClF}}_{\text{3}}$ to make equal $\text{Cl}$ on each side.

  ${\text{FeCl}}_2\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to {\text{FeF}}_3\left(s\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Place 2 as coefficient of ${\text{FeF}}_3$ to make equal $\text{F}$ on each side.

  ${\text{FeCl}}_2\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_3\left(s\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Place 2 as coefficient of ${\text{FeCl}}_2$ to make equal $\text{Fe}$ on each side and change coefficient of ${\text{Cl}}_{\text{2}}$ from 2 to 3.

  ${\text{2FeCl}}_2\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_3\left(s\right)+3{\text{Cl}}_{\text{2}}\left(g\right)$

Since above equation has equal atoms on right and left side so the equation is now balanced.