Chapter 8, Problem 10PE

(a)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

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

Answer to Problem 10PE

The balanced equation is,

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

Explanation of Solution

The given equation is,

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

Place the coefficient 2 in front of ${\text{SO}}_{\text{2}}$ to balance both sulfur and oxygen atoms.

Place the coefficient 2 in front of ${\text{SO}}_{\text{3}}$ to balance both sulfur and oxygen atoms.  The balanced equation is,

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

Each side has 2 sulfur atoms and six oxygen atoms.

(b)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{Li}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{LiOH}\left(\text{aq}\right)$

Answer to Problem 10PE

The balanced equation is,

  ${\text{Li}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{2LiOH}\left(\text{aq}\right)$

Explanation of Solution

The given equation is,

  ${\text{Li}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{LiOH}\left(\text{aq}\right)$

Lithium, hydrogen and oxygen atoms are unbalanced.

Place the coefficient 2 before $\text{LiOH}$ to balance lithium, hydrogen and oxygen atoms.

The balanced equation is,

  ${\text{Li}}_{\text{2}}\text{O}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{2LiOH}\left(\text{aq}\right)$

Each side has two lithium atoms, two oxygen atoms and two hydrogen atoms.

(c)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

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

Answer to Problem 10PE

The balanced equation is,

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

Explanation of Solution

The given equation is,

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

All the atoms are unbalanced.

Place the coefficient 2 before $\text{Na}$ to balance sodium atoms and place the coefficient 2 before $\text{NaOH}$ to balance sodium, hydrogen and oxygen atoms.

Place the coefficient 2 before ${\text{H}}_{\text{2}}\text{O}$ to balance oxygen atoms and hydrogen atoms.

The balanced equation is,

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

Each side has two sodium atoms, four atoms of hydrogen and two oxygen atoms.

(d)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{AgNO}}_{\text{3}}\text{+Ni}\stackrel{}{\to }\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{+Ag}$

Answer to Problem 10PE

The balanced equation is,

  ${\text{2AgNO}}_{\text{3}}\text{+Ni}\stackrel{}{\to }\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{+2Ag}$

Explanation of Solution

The given equation is,

  ${\text{AgNO}}_{\text{3}}\text{+Ni}\stackrel{}{\to }\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{+Ag}$

Treat the unbalanced ${\text{NO}}_{\text{3}}{}^{\text{-}}$ ion as a unit and it is balanced by placing a coefficient by 2 in front of ${\text{AgNO}}_{\text{3}}\text{.}$

Silver atom is balanced by placing a coefficient 2 in front of $\text{Ag}\text{.}$

The balanced equation is,

  ${\text{2AgNO}}_{\text{3}}\text{+Ni}\stackrel{}{\to }\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{+2Ag}$

Each side has two silver atoms, one nickel atom and two ${\text{NO}}_{\text{3}}{}^{\text{-}}$ ions.

(e)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\text{+HCl}\stackrel{}{\to }{\text{BiCl}}_{\text{3}}{\text{+H}}_{\text{2}}\text{S}$

Answer to Problem 10PE

The balanced equation is,

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\text{+6HCl}\stackrel{}{\to }{\text{2BiCl}}_{\text{3}}{\text{+3H}}_{\text{2}}\text{S}$

Explanation of Solution

The given equation is,

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\text{+HCl}\stackrel{}{\to }{\text{BiCl}}_{\text{3}}{\text{+H}}_{\text{2}}\text{S}$

Place the coefficient 6 before $\text{HCl}$ to balance hydrogen atoms and place the coefficient 3 before ${\text{H}}_{\text{2}}\text{S}$ to balance hydrogen atoms.

Place the coefficient 6 before $\text{HCl}$ to balance chlorine atoms and place the coefficient 2 before ${\text{BiCl}}_{\text{3}}$ to balance chlorine atoms.

Place the coefficient 2 before ${\text{BiCl}}_{\text{3}}$ to balance bismuth atoms.

Place the coefficient 3 before ${\text{H}}_{\text{2}}\text{S}$ to balance sulfur atoms.

The balanced equation is,

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\text{+6HCl}\stackrel{}{\to }{\text{2BiCl}}_{\text{3}}{\text{+3H}}_{\text{2}}\text{S}$

Each side has 2 bismuth atoms, 3 sulfur atoms, 6 hydrogen atoms and 6 chlorine atoms.

(f)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{PbO}}_{\text{2}}\stackrel{}{\to }{\text{PbO+O}}_{\text{2}}$

Answer to Problem 10PE

The balanced equation is,

  ${\text{2PbO}}_{\text{2}}\stackrel{}{\to }{\text{2PbO+O}}_{\text{2}}$

Explanation of Solution

The given equation is,

  ${\text{PbO}}_{\text{2}}\stackrel{}{\to }{\text{PbO+O}}_{\text{2}}$

Place the coefficient 2 before ${\text{PbO}}_{\text{2}}$ to balance oxygen atoms and place the coefficient 2 before $\text{PbO}$ to balance oxygen atoms.

The atoms of lead get balanced while balancing oxygen atoms.

The balanced equation is,

  ${\text{2PbO}}_{\text{2}}\stackrel{}{\to }{\text{2PbO+O}}_{\text{2}}$

Each side has 2 lead atoms and four oxygen atoms.

(g)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\left(\text{aq}\right)\text{+KCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{+KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

Answer to Problem 10PE

The balanced equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\left(\text{aq}\right)\text{+2KCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{+2KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

Explanation of Solution

The given equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\left(\text{aq}\right)\text{+KCl}\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{+KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

Place the coefficient 2 before $\text{KCl}$ to balance potassium and chlorine atoms.

Place the coefficient 2 before ${\text{KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ to carbon, potassium, hydrogen and oxygen atoms.

The balanced equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\left(\text{aq}\right)\text{+2KCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{+2KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

Each side has two mercury atoms, four carbon atoms, six hydrogen atoms, four oxygen atoms, two potassium atoms and two chlorine atoms.

(h)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{KI+Br}}_{\text{2}}\stackrel{}{\to }{\text{KBr+I}}_2$

Answer to Problem 10PE

The balanced equation is,

  ${\text{2KI+Br}}_{\text{2}}\stackrel{}{\to }{\text{2KBr+I}}_{\text{2}}$

Explanation of Solution

The given equation is,

  ${\text{KI+Br}}_{\text{2}}\stackrel{}{\to }{\text{KBr+I}}_2$

Place the coefficient 2 before $\text{KI}$ to balance potassium atoms and iodine atoms.

Place the coefficient 2 before $\text{KBr}$ to balance potassium atoms and bromine atoms.

The balanced equation is,

  ${\text{2KI+Br}}_{\text{2}}\stackrel{}{\to }{\text{2KBr+I}}_{\text{2}}$

Each side has 2 potassium atoms, 2 bromine atoms, and two iodine atoms.

(i)

Interpretation Introduction

Interpretation:

The given equation has to be balanced.

The equation is,

  ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+BaCl}}_{\text{2}}\stackrel{}{\to }{\text{KCl+Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

Answer to Problem 10PE

The balanced equation is,

  ${\text{2K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+3BaCl}}_{\text{2}}\stackrel{}{\to }{\text{6KCl+Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

Explanation of Solution

The given equation is,

  ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+BaCl}}_{\text{2}}\stackrel{}{\to }{\text{KCl+Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

Treat the unbalanced ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ as one unit and it is balanced by placing a coefficient 2 before ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}\text{.}$

Barium and chlorine atoms are balanced by placing the coefficient 3 before ${\text{BaCl}}_{\text{2}}.$

Potassium and chlorine atoms are balanced by placing the coefficient 6 before $\text{KCl}\text{.}$

The balanced equation is,

  ${\text{2K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+3BaCl}}_{\text{2}}\stackrel{}{\to }{\text{6KCl+Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

Each side has six potassium atoms, three barium atoms, six chlorine atoms and two ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ ions.