Q4.120CP - Chemistry: Molecular Nature Of Matter And Change

Question

For the following aqueous reactions, complete and balance the molecular equation and write a net ionic equation:

(a) Manganese(II) sulfide + hydrobromic acid

(b) Potassium carbonate + strontium nitrate

(c) Potassium nitrite + hydrochloric acid

(d) Calcium hydroxide + nitric acid

(e) Barium acetate + iron(II) sulfate

(f) Zinc carbonate + sulfuric acid

(g) Copper(II) nitrate + hydrosulfuric acid

(h) Magnesium hydroxide + chloric acid

(i) Potassium chloride + ammonium phosphate

(j) Barium hydroxide + hydrocyanic acid

Step-by-Step Solution

Answer

The balanced equations are produced by the given reactions.

Step 1: Meaning of molecular and net ionic equation

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.

Step 2: Reaction (a) Manganese(II) sulfide + hydrobromic acid

The molecular equation for the given reaction is,

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

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

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

 

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

 

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

Step 3: Reaction (b) Potassium carbonate + strontium nitrate

The molecular equation for the given reaction is,

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

 

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

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

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

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

Step 4: Reaction (c) Potassium nitrite + hydrochloric acid

The molecular equation for the given reaction is,

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

 

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

Step 5: Reaction (d) Calcium hydroxide + nitric acid

The molecular equation for the given reaction is,

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

 

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

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

 

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

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

 

Step 6: Reaction (e) Barium acetate + iron(II) sulfate

The molecular equation for the given reaction is,

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

 

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

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

 

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

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

Step 7: Reaction (f) Zinc carbonate + sulfuric acid

The molecular equation for the given reaction is,

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

 

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

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

 

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

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

Step 8: Reaction (g) Copper(II) nitrate + hydrosulfuric acid

The molecular equation for the given reaction is,

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

 

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

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

 

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

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

Step 9: Reaction (h) Magnesium hydroxide + chloric acid

The molecular equation for the given reaction is,

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

 

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

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

 

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

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

Step 10: Reaction (i) Potassium chloride + ammonium phosphate

The molecular equation for the given reaction is,

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

 

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

 

Step 11: Reaction (j) Barium hydroxide + hydrocyanic acid

The molecular equation for the given reaction is,

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

 

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

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

 

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

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