Chapter 3.4, Problem 3.10CE

(a)

Interpretation Introduction

Interpretation:

The product has to be predicted, balanced and net ionic reaction for the reaction of ${\text{BaCl}}_{\text{2}}{\text{(aq) and Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq)}$ has to be written.

Concept introduction:

Balancing reaction:

Balanced reaction is a chemical reaction in which number of atoms for each element in the reaction and the total charge are same on both reactant side and the product side.

Steps in balancing the information

• Step 1: Write the unbalanced equation
• Step 2: Find the coefficient to balance the equation.
• The coefficient should be reduced to the smallest whole number.

Gas forming reaction: The reaction of acid and metal carbonates which produce carbonic acid. The carbonic acid decomposes which gives water and carbon dioxide.

Most of the ionic compounds are soluble in water, very few of the ionic compounds are sparingly soluble, and some of the ionic compounds are insoluble in water. When it is soluble in water ions gets separated in the solution.

Explanation of Solution

The given reactant is shown below,

  ${\text{BaCl}}_{\text{2}}{\text{(aq) and Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq)}$

The product of the reaction is shown below,

  ${\text{BaCl}}_{\text{2}}{\text{(aq) + Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq) }\to {\text{ BaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s) + 2NaCl(aq)}$

Product of the reaction is barium oxalate and sodium chloride.

Balance the equation,

    ${\text{BaCl}}_{\text{2}}{\text{(aq) + Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq) }\to {\text{ BaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s) + NaCl(aq)}$

Balance the equation, when balancing the equation, we should not alter the subscripts and we can change coefficients. There are two sodium atoms in left side of the reaction and one sodium atom in the right of the reaction. Therefore, two molecule of sodium chloride is added to the right side of the reaction. Therefore, the balanced equation is given below.

      ${\text{BaCl}}_{\text{2}}{\text{(aq) + Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq) }\to {\text{ BaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s) + 2NaCl(aq)}$

The complete ionic equation is given below,

    ${\text{Ba}}^{\text{2+}}{\text{(aq) + 2Cl}}^{\text{- }}\text{(aq) }+{\text{2Na}}^{\text{+}}{\text{(aq) + C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{BaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s)}+{\text{2Na}}^{\text{+}}{\text{(aq) + 2Cl}}^{\text{- }}\text{(aq)}$

Net ionic equation of the given reaction shown below

${\text{Ba}}^{\text{2+}}{\text{(aq) + C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{BaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s)}$

(b)

Interpretation Introduction

Interpretation:

The product has to be predicted, balanced and net ionic reaction for the reaction of ${\text{Sr(OH)}}_{\text{2}}\text{(s)}{\text{and HNO}}_{\text{3}}\text{(aq)}$ has to be written.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given reactant is shown below,

  ${\text{Sr(OH)}}_{\text{2}}\text{(s)}{\text{and HNO}}_{\text{3}}\text{(aq)}$

The product of the reaction is shown below,

  ${\text{2 HNO}}_{\text{3}}{\text{(aq) + Sr(OH)}}_{\text{2}}\text{(s) }\to {\text{ Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

Product of the reaction is strontium nitrate and water.

Balance the equation,

    ${\text{HNO}}_{\text{3}}{\text{(aq) + Sr(OH)}}_{\text{2}}\text{(s) }\to {\text{ Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

Balance the equation, when balancing the equation, we should not alter the subscripts and we can change coefficients. There are two hydrogen atoms in the right side and one hydrogen atom in the left side. Therefore, two molecule of nitric acid is added to left side of reaction. Now, the balanced equation is given below.

  ${\text{2 HNO}}_{\text{3}}{\text{(aq) + Sr(OH)}}_{\text{2}}\text{(s) }\to {\text{ Sr(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

The complete ionic equation is given below,

    ${\text{ 2H}}^{\text{+}}{\text{(aq) + 2 NO}}_3^{2-}{\text{(aq) + Sr(OH)}}_{\text{2}}\text{(s) }\to {\text{ Sr}}^{2+}{\text{(aq)+ NO}}_3^{2-}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

Net ionic equation of the given reaction shown below

  ${\text{ 2H}}^{\text{+}}{\text{(aq) + Sr(OH)}}_{\text{2}}\text{(s) }\to {\text{ Sr}}^{2+}{\text{(aq)+ H}}_{\text{2}}\text{O(l)}$

(c)

Interpretation Introduction

Interpretation:

The product has to be predicted, balanced and net ionic reaction for the reaction of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) and NiCl}}_{\text{2}}\text{(aq)}$ has to be written.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given reactant is shown below,

  ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) and NiCl}}_{\text{2}}\text{(aq)}$

The product of the reaction is shown below,

  ${\text{2(NH}}_{\text{4}}{\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3NiCl}}_{\text{2}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}{\text{(s) + 6NH}}_{\text{4}}\text{Cl(aq)}$

Product of the reaction is nickel(II) phosphate and aqueous ammonium chloride.

Balance the equation,

    ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + NiCl}}_{\text{2}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}{\text{(s) + NH}}_{\text{4}}\text{Cl(aq)}$

Balance the equation, when balancing the equation, we should not alter the subscripts and we can change coefficients. There are three nickel atoms in the right side and one nickel atom in the left side. Therefore, three molecule of nickel chloride is added to left side of reaction. Now, the balanced equation is given below.

  ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3NiCl}}_{\text{2}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}{\text{(s) + NH}}_{\text{4}}\text{Cl(aq)}$

There are six chlorine atoms in the left side and one chlorine atom in the right side. Therefore, six molecule of ammonium chloride is added to right side of reaction. Now, the balanced equation is given below.

  ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3NiCl}}_{\text{2}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}{\text{(s) + 6NH}}_{\text{4}}\text{Cl(aq)}$

There are two phosphorus atoms in the right side and one phosphorus atom in the left side. Therefore, two molecule of ammonium phosphate is added to left side of reaction. Now, the balanced equation is given below.

  ${\text{2(NH}}_{\text{4}}{\text{)}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3NiCl}}_{\text{2}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}{\text{(s) + 6NH}}_{\text{4}}\text{Cl(aq)}$

The complete ionic equation is given below,

    ${\text{6NH}}_4^{+}{\text{(aq) + 2PO}}_4^{3-}{\text{(aq) + 3Ni}}^{\text{2+}}{\text{(aq) + 6Cl}}^{\text{-}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}{\text{(s) + 6NH}}_4^{+}{\text{(aq) + 6Cl}}^{\text{-}}\text{(aq)}$

Net ionic equation of the given reaction shown below

  ${\text{2PO}}_4^{3-}{\text{(aq) + 3Ni}}^{\text{2+}}\text{(aq) }\to {\text{ Ni}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s) }$

(d)

Interpretation Introduction

Interpretation:

The product has to be predicted, balanced and net ionic reaction for the reaction of ${\text{KOH (aq) and HClO}}_{\text{4}}\text{(aq)}$ has to be written.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given reactant is shown below,

  ${\text{KOH (aq) and HClO}}_{\text{4}}\text{(aq)}$

The product of the reaction is shown below,

  ${\text{KOH(aq) + HClO}}_{\text{4}}\text{(aq) }\to {\text{ H}}_{\text{2}}{\text{O(l) + KClO}}_{\text{4}}\text{(aq)}$

Product of the reaction is aqueous potassium chlorate and water.

Balance the equation,

    ${\text{KOH(aq) + HClO}}_{\text{4}}\text{(aq) }\to {\text{ H}}_{\text{2}}{\text{O(l) + KClO}}_{\text{4}}\text{(aq)}$

Balance the equation, when balancing the equation, we should not alter the subscripts and we can change coefficients. The reaction is balanced. Therefore, the balanced equation is given below.

  ${\text{KOH(aq) + HClO}}_{\text{4}}\text{(aq) }\to {\text{ H}}_{\text{2}}{\text{O(l) + KClO}}_{\text{4}}\text{(aq)}$

The complete ionic equation is given below,

    ${\text{K}}^{\text{+}}{\text{(aq) + OH}}^{-}{\text{(aq) + H}}^{\text{+}}{\text{(aq) + ClO}}_4^{-}\text{(aq) }\to {\text{ H}}_{\text{2}}{\text{O(l) + K}}^{\text{+}}\text{(aq)}+{\text{ClO}}_4^{-}\text{(aq)}$

Net ionic equation of the given reaction shown below

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