91Ó°ÊÓ

Molecular equation: The chemical equation which is balanced and the ionic compounds are shown as molecules.

Net ionic equation: The chemical equation which is balanced and shows only chemical species that participate in the reaction. The ions that are present in the same form on both sides of the reaction are known as Spectator ions.

The molecular equation for the given reaction is,

$\mathbf{MnS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{HBr}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{MnBr}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

$\mathbf{MnS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{Br}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Mn}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{Br}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

$\mathbf{MnS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Mn}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The molecular equation for the given reaction is,

${\mathbf{K}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Sr}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{2}{\mathbf{KNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{SrCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

$\begin{array}{l}\mathbf{2}{\mathbf{K}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Sr}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{2}{\mathbf{K}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{SrCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\\ \end{array}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

${\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Sr}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{→}{\mathbf{SrCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The molecular equation for the given reaction is,

${\mathbf{KNO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{HCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{KCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{HNO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Both potassium chloride and nitrous acid are soluble. Hence, there will be not net ionic equation.

The molecular equation for the given reaction is,

$\mathbf{Ca}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{Ca}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

${\mathbf{Ca}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{OH}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Ca}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

$\mathbf{2}{\mathbf{OH}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The molecular equation for the given reaction is,

$\mathbf{Ba}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COO}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{FeSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{Fe}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COO}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{BaSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

${\mathbf{Ba}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{COO}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Fe}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{COO}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{BaSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

${\mathbf{Ba}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{BaSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The molecular equation for the given reaction is,

${\mathbf{ZnCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{ZnSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

${\mathbf{ZnCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

${\mathbf{ZnCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The molecular equation for the given reaction is,

$\mathbf{Cu}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{CuS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

${\mathbf{Cu}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{S}}^{\mathbf{−}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{CuS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

${\mathbf{Cu}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{S}}^{\mathbf{−}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{CuS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

The molecular equation for the given reaction is,

$\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{HClO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{Mg}{\mathbf{\left(}{\mathbf{ClO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

$\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{ClO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Mg}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{ClO}}_{\mathbf{3}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

$\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Mg}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The molecular equation for the given reaction is,

$\mathbf{3}\mathbf{KCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{K}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Both tripotassium phosphate and ammonium chloride are soluble. Hence, there will be not net ionic equation.

The molecular equation for the given reaction is,

$\mathbf{Ba}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{HCN}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{Ba}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The total ionic equation can be written as by writing ionic compounds in the form of ions. So, the total ionic equation is,

${\mathbf{Ba}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{OH}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CN}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}{\mathbf{Ba}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CN}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

The net ionic equation can be written as by cancelling the spectator ions. Now, the net ionic equation is,

$\mathbf{2}{\mathbf{OH}}^{\mathbf{−}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{→}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$