Chapter 2, Problem 2.97QP

Balance the following equations.

1. a Sn + NaOH → Na₂SnO₂ + H₂
2. b Al + Fe₃O₄ → Al₂O₃ + Fe
3. c CH₃OH + O₂ → CO₂ + H₂O
4. d P₄O₁₁₁ + H₂O  → H₃PO₄
5. e PCl₅ + H₂O → H₃PO₄ + HCl

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 2.97QP

The balanced equation is given as,

$\text{Sn}\text{+}\text{2NaOH}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{SnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

$\text{Sn}\text{+}\text{NaOH}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{SnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

Start by balancing Sodium.  This can be done by adding coefficient “2” to sodium hydroxide in the reactant side.  By doing this we obtain, the balanced chemical equation as shown below,

$\text{Sn}\text{+}\text{2NaOH}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{SnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

Conclusion

The balanced equation was given as,

$\text{Sn}\text{+}\text{2NaOH}\stackrel{}{\to }{\text{Na}}_{\text{2}}{\text{SnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

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 2.97QP

The balanced equation is given as,

$\text{8Al}\text{+}{\text{3Fe}}_{\text{3}}{\text{O}}_{\text{4}}\stackrel{}{\to }{\text{4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}\text{9Fe}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

$\text{Al}\text{+}{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\stackrel{}{\to }{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}\text{Fe}$

Start by balancing Oxygen as it occurs only once in the product side.  This can be done by adding coefficient “3” to ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ and “4” to ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$.  By doing this we obtain, the chemical equation as shown below,

$\text{Al}\text{+}3{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\stackrel{}{\to }4{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}\text{Fe}$

Now balance the other atoms, iron and aluminium.  This can be done by adding coefficient “8” to aluminium in reactant side and coefficient “9” to iron in the product side.  The balanced chemical equation can be obtained as,

$\text{8Al}\text{+}{\text{3Fe}}_{\text{3}}{\text{O}}_{\text{4}}\stackrel{}{\to }{\text{4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}\text{9Fe}$

Conclusion

The balanced equation was given as,

$\text{8Al}\text{+}{\text{3Fe}}_{\text{3}}{\text{O}}_{\text{4}}\stackrel{}{\to }{\text{4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}\text{9Fe}$

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 2.97QP

The balanced equation is given as,

${\text{2CH}}_{\text{3}}\text{OH}\text{+}3{\text{O}}_2\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}4{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{CH}}_{\text{3}}\text{OH}\text{+}{\text{O}}_2\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

Start by balancing Hydrogen 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,

${\text{CH}}_{\text{3}}\text{OH}\text{+}{\text{O}}_2\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}2{\text{H}}_{\text{2}}\text{O}$

In order to avoid fractional coefficients for oxygen, multiply the above equation with 2.  Then the equation will be obtained as,

${\text{2CH}}_{\text{3}}\text{OH}\text{+}2{\text{O}}_2\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}4{\text{H}}_{\text{2}}\text{O}$

Now balance the oxygen atom by changing the coefficient “2” to “3” in the reactant side for oxygen.  This balances the whole equation.  The balanced equation can be given as,

${\text{2CH}}_{\text{3}}\text{OH}\text{+}3{\text{O}}_2\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}4{\text{H}}_{\text{2}}\text{O}$

Conclusion

The balanced equation was given as,

${\text{2CH}}_{\text{3}}\text{OH}\text{+}3{\text{O}}_2\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}4{\text{H}}_{\text{2}}\text{O}$

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 2.97QP

The balanced equation is given as,

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+}6{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

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,

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

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,

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+}6{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Conclusion

The balanced equation was given as,

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{+}6{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

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 2.97QP

The balanced equation is given as,

${\text{PCl}}_{\text{5}}\text{+}4{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+\text{5HCl}$

Explanation of Solution

The raw chemical equation given in the problem statement is,

${\text{PCl}}_{\text{5}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+\text{HCl}$

Start by balancing Chlorine as it occurs only once in the product side.  This can be done by adding coefficient “5” to the product side.  By doing this we obtain, the chemical equation as shown below,

${\text{PCl}}_{\text{5}}\text{+}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+5\text{HCl}$

Now balance the hydrogen atom by adding the coefficient to “4” to water in the reactant side.  This balances the whole equation.  The balanced equation can be given as,

${\text{PCl}}_{\text{5}}\text{+}4{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+\text{5HCl}$

Conclusion

The balanced equation was given as,

${\text{PCl}}_{\text{5}}\text{+}4{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+\text{5HCl}$