Chapter 2, Problem 2.125QP

(a)

Interpretation Introduction

Interpretation:

The given chemical equation has to be balanced.

Concept Introduction:

When a chemical reaction occurs, the total number of atoms in the reactant has to be same as that in the product formed.  Chemical reaction can be entered in form of chemical equation with all necessary conditions.  The chemical equation has to be balanced by adding coefficients only before the elements or compounds to make the atoms on both sides of arrow equal.  This is known as balanced chemical equation.

Answer to Problem 2.125QP

The balanced equation is given as,

${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{+}7{\text{O}}_{\text{2}}\stackrel{}{\to }4{\text{CO}}_{\text{2}}\text{+}6{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

The fractional coefficient for oxygen can be avoided by doubling all coefficient except oxygen molecule in the left side.  Then the oxygen atom can be balanced by adding coefficient “7” to oxygen molecule in the reactant side.  The balanced chemical equation can be obtained as,

${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{+}7{\text{O}}_{\text{2}}\stackrel{}{\to }4{\text{CO}}_{\text{2}}\text{+}6{\text{H}}_{\text{2}}\text{O}$

Conclusion

The balanced equation was given as,

${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{+}7{\text{O}}_{\text{2}}\stackrel{}{\to }4{\text{CO}}_{\text{2}}\text{+}6{\text{H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

The given chemical equation has to be balanced.

Concept Introduction:

When a chemical reaction occurs, the total number of atoms in the reactant has to be same as that in the product formed.  Chemical reaction can be entered in form of chemical equation with all necessary conditions.  The chemical equation has to be balanced by adding coefficients only before the elements or compounds to make the atoms on both sides of arrow equal.  This is known as balanced chemical equation.

Answer to Problem 2.125QP

The balanced equation is given as,

${\text{P}}_{\text{4}}{\text{O}}_6\text{+}6{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_3$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{P}}_{\text{4}}{\text{O}}_6\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_3$

Start by balancing Phosphorous as it occurs only once in the product side.  This can be done by adding coefficient “4” to the product side.  By doing this we obtain, the chemical equation as shown below,

${\text{P}}_{\text{4}}{\text{O}}_6\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_3$

Now balance the hydrogen atom by adding the coefficient to “6” to water in the reactant side.  This balances the whole equation.  The balanced equation can be given as,

${\text{P}}_{\text{4}}{\text{O}}_6\text{+}6{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_3$

Conclusion

The balanced equation was given as,

${\text{P}}_{\text{4}}{\text{O}}_6\text{+}6{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_3$

(c)

Interpretation Introduction

Interpretation:

The given chemical equation has to be balanced.

Concept Introduction:

When a chemical reaction occurs, the total number of atoms in the reactant has to be same as that in the product formed.  Chemical reaction can be entered in form of chemical equation with all necessary conditions.  The chemical equation has to be balanced by adding coefficients only before the elements or compounds to make the atoms on both sides of arrow equal.  This is known as balanced chemical equation.

Answer to Problem 2.125QP

The balanced equation is given as,

${\text{4KClO}}_{\text{3}}\stackrel{}{\to }\text{KCl}\text{+}{\text{3KClO}}_{\text{4}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{KClO}}_{\text{3}}\stackrel{}{\to }\text{KCl}\text{+}{\text{KClO}}_{\text{4}}$

Start by balancing oxygen as it occurs only once in the product side.  This can be done by adding coefficient “3” to ${\text{KClO}}_{\text{4}}$ and coefficient “4” to ${\text{KClO}}_3$.  By doing this we obtain, the balanced chemical equation as shown below,

${\text{4KClO}}_{\text{3}}\stackrel{}{\to }\text{KCl}\text{+}3{\text{KClO}}_{\text{4}}$

Conclusion

The balanced equation was given as,

${\text{4KClO}}_{\text{3}}\stackrel{}{\to }\text{KCl}\text{+}3{\text{KClO}}_{\text{4}}$

(d)

Interpretation Introduction

Interpretation:

The given chemical equation has to be balanced.

Concept Introduction:

When a chemical reaction occurs, the total number of atoms in the reactant has to be same as that in the product formed.  Chemical reaction can be entered in form of chemical equation with all necessary conditions.  The chemical equation has to be balanced by adding coefficients only before the elements or compounds to make the atoms on both sides of arrow equal.  This is known as balanced chemical equation.

Answer to Problem 2.125QP

The balanced equation is given as,

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}\text{2NaOH}\stackrel{}{\to }{\text{2NH}}_{\text{3}}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}\text{NaOH}\stackrel{}{\to }{\text{NH}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Start by balancing nitrogen as it occurs only once in the product side.  This can be done by adding coefficient “2” to ${\text{NH}}_{\text{3}}$ in the product side.  By doing this we obtain, the chemical equation as shown below,

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}\text{NaOH}\stackrel{}{\to }{\text{2NH}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Now balance the sodium and oxygen atom by adding the coefficient to “2” to water in the product side and sodium hydroxide in the reactant side.  This balances the whole equation.  The balanced equation can be given as,

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}\text{2NaOH}\stackrel{}{\to }{\text{2NH}}_{\text{3}}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Conclusion

The balanced equation was given as,

${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}\text{2NaOH}\stackrel{}{\to }{\text{2NH}}_{\text{3}}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

(e)

Interpretation Introduction

Interpretation:

The given chemical equation has to be balanced.

Concept Introduction:

When a chemical reaction occurs, the total number of atoms in the reactant has to be same as that in the product formed.  Chemical reaction can be entered in form of chemical equation with all necessary conditions.  The chemical equation has to be balanced by adding coefficients only before the elements or compounds to make the atoms on both sides of arrow equal.  This is known as balanced chemical equation.

Answer to Problem 2.125QP

The balanced equation is given as,

${\text{2NBr}}_{\text{3}}\text{+}\text{3NaOH}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{+}\text{3NaBr}\text{+}\text{3HOBr}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{NBr}}_{\text{3}}\text{+}\text{NaOH}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{+}\text{NaBr}\text{+}\text{HOBr}$

Start by balancing nitrogen as it occurs only once in the product side.  This can be done by adding coefficient “2” to nitrogen tribromide in the reactant side.  By doing this we obtain, the chemical equation as shown below,

${\text{2NBr}}_{\text{3}}\text{+}\text{NaOH}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{+}\text{NaBr}\text{+}\text{HOBr}$

Now balance the other atoms by adding the coefficient to “3” to all the compounds except nitrogen.  This balances the whole equation.  The balanced equation can be given as,

${\text{2NBr}}_{\text{3}}\text{+}\text{3NaOH}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{+}\text{3NaBr}\text{+}\text{3HOBr}$

Conclusion

The balanced equation was given as,

${\text{2NBr}}_{\text{3}}\text{+}\text{3NaOH}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{+}\text{3NaBr}\text{+}\text{3HOBr}$