Chapter 2, Problem 2.126QP

Balance the following equations.

1. a NaOH + H₂CO₃ → Na₂CO₃ + H₂O
2. b SiCl₄ + H₂O → SiO₂ + HCl
3. c Ca₃(PO₄)₂ + C → Ca₃P₂ + CO
4. d H₂S + O₂ → SO₂ + H₂O
5. e N₂O₅ → NO₂ → O₂

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 2.126QP

The balanced equation is given as,

$\text{2NaOH}\text{+}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}2{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

$\text{NaOH}\text{+}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}\text{O}$

Start balancing sodium as this occurs only once in the product side.  This can be done by adding coefficient “2” to sodium hydroxide in the reactant side.  The equation is obtained as,

$\text{2NaOH}\text{+}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}{\text{H}}_{\text{2}}\text{O}$

Now balance hydrogen atom on both side.  This can be done by adding coefficient “2” to water in the product side.  This balanced the oxygen atom also.  Hence, the balanced chemical equation obtained can be given as,

$\text{2NaOH}\text{+}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}2{\text{H}}_{\text{2}}\text{O}$

Conclusion

The balanced equation was given as,

$\text{2NaOH}\text{+}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}2{\text{H}}_{\text{2}}\text{O}$

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 2.126QP

The balanced equation is given as,

${\text{SiCl}}_{\text{4}}\text{+}{\text{2H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{SiO}}_{\text{2}}\text{+}\text{4HCl}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{SiCl}}_{\text{4}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{SiO}}_{\text{2}}\text{+}\text{HCl}$

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

${\text{SiCl}}_{\text{4}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{SiO}}_{\text{2}}\text{+}4\text{HCl}$

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

${\text{SiCl}}_{\text{4}}\text{+}2{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{SiO}}_{\text{2}}\text{+}4\text{HCl}$

Conclusion

The balanced equation was given as,

${\text{SiCl}}_{\text{4}}\text{+}{\text{2H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{SiO}}_{\text{2}}\text{+}\text{4HCl}$

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 2.126QP

The balanced equation is given as,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}\text{8C}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\text{+}\text{8CO}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

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

Start by balancing oxygen as it occurs only once in the product side.  This can be done by adding coefficient “8” to $\text{CO}$.  By doing this we obtain, the chemical equation as shown below,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}\text{C}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\text{+}8\text{CO}$

Now start balancing carbon atom.  This can be done by adding coefficient “8” to $\text{C}$.  By doing this we obtain, the balanced chemical equation as shown below,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}8\text{C}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\text{+}8\text{CO}$

Conclusion

The balanced equation was given as,

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{+}\text{8C}\stackrel{}{\to }{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\text{+}\text{8CO}$

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 2.126QP

The balanced equation is given as,

${\text{2H}}_{\text{2}}\text{S}\text{+}{\text{3O}}_{\text{2}}\stackrel{}{\to }{\text{2SO}}_{\text{2}}\text{+}{\text{2H}}_{\text{2}}\text{O}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{H}}_{\text{2}}\text{S}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{SO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

Start by balancing oxygen atom by adding coefficient “3” to oxygen molecule in the reactant side and coefficient “2” to the substance present in the product side.  By doing this we obtain, the chemical equation as shown below,

${\text{H}}_{\text{2}}\text{S}\text{+}{\text{3O}}_{\text{2}}\stackrel{}{\to }{\text{2SO}}_{\text{2}}\text{+}{\text{2H}}_{\text{2}}\text{O}$

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

${\text{2H}}_{\text{2}}\text{S}\text{+}{\text{3O}}_{\text{2}}\stackrel{}{\to }{\text{2SO}}_{\text{2}}\text{+}{\text{2H}}_{\text{2}}\text{O}$

Conclusion

The balanced equation was given as,

${\text{2H}}_{\text{2}}\text{S}\text{+}{\text{3O}}_{\text{2}}\stackrel{}{\to }{\text{2SO}}_{\text{2}}\text{+}{\text{2H}}_{\text{2}}\text{O}$

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 2.126QP

The balanced equation is given as,

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }{\text{4NO}}_{\text{2}}\text{+}{\text{O}}_{\text{2}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }{\text{NO}}_{\text{2}}\text{+}{\text{O}}_{\text{2}}$

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

${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }2{\text{NO}}_{\text{2}}\text{+}{\text{O}}_{\text{2}}$

Now balance the oxygen atom.  Totally there are five oxygen atoms in the reactant side, while in the product side there are six oxygen atoms.  Multiply by ½ in front of Oxygen molecule in the product side gives the equation as,

${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }2{\text{NO}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}$

Multiply the whole equation by 2.  This gives the balanced equation as,

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }4{\text{NO}}_{\text{2}}\text{+}{\text{O}}_{\text{2}}$

Conclusion

The balanced equation was given as,

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\stackrel{}{\to }{\text{4NO}}_{\text{2}}\text{+}{\text{O}}_{\text{2}}$