Chapter 2, Problem 2.98QP

Balance the following equations.

1. a Ca₃(PO₄)₂ + H₃PO₄ → Ca(H₂PO₄)₂
2. b MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
3. c Na₂S₂O₃ + I₂ → NaI + Na₂S₄O₆
4. d Al₄C₃ + H₂O → Al(OH)₃ + CH₄
5. e NO₂ + H₂O → HNO₃ + NO

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 2.98QP

The balanced equation is given as,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}{\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}\stackrel{}{\to }{\text{3Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\stackrel{}{\to }{\text{Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

Start by balancing Hydrogen.  This can be done by adding coefficient “3” to the reactant side.  By doing this we obtain, the balanced chemical equation as shown below,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\stackrel{}{\to }{\text{3Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

Finally, balance phosphorus by adding coefficient “4” in front of phosphoric acid in the reactant side.  The balanced equation can be obtained as,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\stackrel{}{\to }{\text{3Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

Conclusion

The balanced equation was given as,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}{\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}\stackrel{}{\to }{\text{3Ca(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

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 2.98QP

The balanced equation is given as,

${\text{MnO}}_{\text{2}}\text{+}\text{4HCl}\stackrel{}{\to }{\text{MnCl}}_{\text{2}}\text{+}{\text{Cl}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{MnO}}_{\text{2}}\text{+}\text{HCl}\stackrel{}{\to }{\text{MnCl}}_{\text{2}}\text{+}{\text{Cl}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

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

${\text{MnO}}_{\text{2}}\text{+}\text{HCl}\stackrel{}{\to }{\text{MnCl}}_{\text{2}}\text{+}{\text{Cl}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

Now balance the other atoms, hydrogen and chlorine.  This can be done by adding coefficient “4” to hydrogen chloride in reactant side.  The balanced chemical equation can be obtained as,

${\text{MnO}}_{\text{2}}\text{+}\text{4HCl}\stackrel{}{\to }{\text{MnCl}}_{\text{2}}\text{+}{\text{Cl}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$

Conclusion

The balanced equation was given as,

${\text{MnO}}_{\text{2}}\text{+}\text{4HCl}\stackrel{}{\to }{\text{MnCl}}_{\text{2}}\text{+}{\text{Cl}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$

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 2.98QP

The balanced equation is given as,

${\text{2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{I}}_{\text{2}}\stackrel{}{\to }2\text{NaI}\text{+}{\text{Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{I}}_{\text{2}}\stackrel{}{\to }\text{NaI}\text{+}{\text{Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$

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

${\text{2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{I}}_{\text{2}}\stackrel{}{\to }\text{NaI}\text{+}{\text{Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$

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

${\text{2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{I}}_{\text{2}}\stackrel{}{\to }2\text{NaI}\text{+}{\text{Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$

Conclusion

The balanced equation was given as,

${\text{2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{I}}_{\text{2}}\stackrel{}{\to }2\text{NaI}\text{+}{\text{Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$

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 2.98QP

The balanced equation is given as,

${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\text{+}12{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{Al(OH)}}_{\text{3}}+{\text{3CH}}_{\text{4}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{Al(OH)}}_{\text{3}}+{\text{CH}}_{\text{4}}$

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

${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{Al(OH)}}_{\text{3}}+{\text{CH}}_{\text{4}}$

Now balance the hydrogen and oxygen atom by adding the coefficient to “12” to water in the reactant side.  This balances the whole equation.  The balanced equation can be given as,

${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\text{+}12{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{Al(OH)}}_{\text{3}}+{\text{3CH}}_{\text{4}}$

Conclusion

The balanced equation was given as,

${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\text{+}12{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{Al(OH)}}_{\text{3}}+{\text{3CH}}_{\text{4}}$

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 2.98QP

The balanced equation is given as,

${\text{3NO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }2{\text{HNO}}_{\text{3}}+\text{NO}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{NO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{HNO}}_{\text{3}}+\text{NO}$

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

${\text{NO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }2{\text{HNO}}_{\text{3}}+\text{NO}$

Now balance the nitrogen atom by adding the coefficient to “3” to nitrogen dioxide in the reactant side.  This balances the whole equation.  The balanced equation can be given as,

${\text{3NO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }2{\text{HNO}}_{\text{3}}+\text{NO}$

Conclusion

The balanced equation was given as,

${\text{3NO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }2{\text{HNO}}_{\text{3}}+\text{NO}$