Chapter 3, Problem 63PS

Balance each of the following equations, and classify them as precipitation, acid-base, gas-forming, or oxidation-reduction reactions. Show states for reactants and products (s, ℓ, g, aq).

(a) CuCl₂ + H₂S →CuS + HCl

(b) H₃PO₄ + KOH → H₂O + K₃PO₄

(c) Ca +HBr → H₂ + CaBr₂

(d) MgC1₂ + NaOH → Mg(OH)₂ + NaCl

Interpretation Introduction

Interpretation:

• The given reaction has to be classified as precipitation or acid-base or gas forming reaction and the reaction has to be balanced.
• State of the product and reactant should be shown.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the ions combine in the solution is called precipitation reaction.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

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

Oxidation - reduction reaction: The electrons are transferred to one to other is called oxidation reduction reaction.

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.

Soluble compounds in water

Almost all the salts of ${\text{Na}}^{\text{+}}{\text{, K}}^{\text{+}}{\text{, NH}}_{\text{4}}{}^{\text{+}}\text{,}\left({\text{NO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{4}}{}^{\text{-}}\right)\text{and}\left({\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right)$ are soluble.

Almost all the salts of ${\text{Cl}}^{\text{-}}{\text{, Br}}^{\text{-}}{\text{, I}}^{\text{-}}\left(\text{halides}\right)$ are soluble.  But some of the halides are $\left({\text{Ag}}^{\text{+}}{\text{, Hg}}_{\text{2}}{}^{\text{2+}}{\text{, Pb}}^{\text{2+}}\right)$ insoluble.

Salts of F^- are soluble.  But some of the fluoride salt of ${\text{Mg}}^{\text{2+}}{\text{, Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}$ are insoluble.

Salts of $\text{sulfate }\left({\text{SO}}_{\text{4}}{}^{\text{2-}}\right)$ are soluble.  But sulfates of ${\text{Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}{\text{, Ag}}^{\text{+}}\text{,}$ are insoluble.

Insoluble compounds in water:

Most of the salts of $\left({\text{CO}}_{\text{3}}{}^{\text{2-}}\right)\text{,}\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{, }\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)\text{,}\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)\text{and}\left({\text{S}}^{\text{2-}}\right)$ are insoluble but some of the salts of NH₄⁺, alkali metal cations and BaS are soluble.

Most of the metal hydroxides and oxides are insoluble in water bit some of the alkali metal hydroxides, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\text{and Sr}{\left(\text{OH}\right)}_{\text{2}}$ are soluble in water.

Answer to Problem 63PS

The given reaction is precipitation reaction and the state of reactant and product are shown below.

     

Explanation of Solution

The balancing of the reaction and the type of the reaction is shown below, the given reaction is precipitation reaction and the state of the reaction is shown below,

     

The given compound is copper chloride and hydrogen sulfide which is soluble in water.  In this reaction copper chloride reaction with hydrogen sulfide to give copper sulfide and hydrochloric acidater.  Most of the salts of $\left({\text{CO}}_{\text{3}}{}^{\text{2-}}\right)\text{,}\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{,}\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)\text{,}\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)\text{and}\left({\text{S}}^{\text{2-}}\right)$ are insoluble but some of the salts of NH₄⁺, alkali metal cations and BaS are soluble.

Therefore the given reaction is precipitation reaction

    

Balance the equation,

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

    

Interpretation Introduction

Interpretation:

• The given reaction has to be classified as precipitation or acid-base or gas forming reaction and the reaction has to be balanced.
• State of the product and reactant should be shown.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the ions combine in the solution is called precipitation reaction.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

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

Oxidation - reduction reaction: The electrons are transferred to one to other is called oxidation reduction reaction.

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.

Soluble compounds in water

Almost all the salts of ${\text{Na}}^{\text{+}}{\text{, K}}^{\text{+}}{\text{, NH}}_{\text{4}}{}^{\text{+}}\text{,}\left({\text{NO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{4}}{}^{\text{-}}\right)\text{and}\left({\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right)$ are soluble.

Almost all the salts of ${\text{Cl}}^{\text{-}}{\text{, Br}}^{\text{-}}{\text{, I}}^{\text{-}}\left(\text{halides}\right)$ are soluble.  But some of the halides are $\left({\text{Ag}}^{\text{+}}{\text{, Hg}}_{\text{2}}{}^{\text{2+}}{\text{, Pb}}^{\text{2+}}\right)$ insoluble.

Salts of F^- are soluble.  But some of the fluoride salt of ${\text{Mg}}^{\text{2+}}{\text{, Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}$ are insoluble.

Salts of $\text{sulfate }\left({\text{SO}}_{\text{4}}{}^{\text{2-}}\right)$ are soluble.  But sulfates of ${\text{Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}{\text{, Ag}}^{\text{+}}\text{,}$ are insoluble.

Insoluble compounds in water:

Most of the salts of $\left({\text{CO}}_{\text{3}}{}^{\text{2-}}\right)\text{,}\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{, }\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)\text{,}\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)\text{and}\left({\text{S}}^{\text{2-}}\right)$ are insoluble but some of the salts of NH₄⁺, alkali metal cations and BaS are soluble.

Most of the metal hydroxides and oxides are insoluble in water bit some of the alkali metal hydroxides, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\text{and Sr}{\left(\text{OH}\right)}_{\text{2}}$ are soluble in water.

Answer to Problem 63PS

The balancing of the reaction and the type of the reaction is shown below, the given reaction is acid - base reaction and the state of the reaction is shown below.

  

Explanation of Solution

The balancing of the reaction and the type of the reaction is shown below, the given reaction is acid - base reaction and the state of the reaction is shown below

  

The given compound is potassium hydroxide and Phosphoric acid.  In this reaction potassium hydroxide reaction with Phosphoric acid to give Tripotassium phosphate and formation of water is also the product.

    

Balance the equation,

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

    

Balance the hydrogen atom in the given equation.  There are six hydrogen atoms in the left side and two hydrogen atoms in the right side.  Therefore three molecule of water is added to right side of reaction.  Therefore the balanced equation is given below.

    

Interpretation Introduction

Interpretation:

• The given reaction has to be classified as precipitation or acid-base or gas forming reaction and the reaction has to be balanced.
• State of the product and reactant should be shown.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the ions combine in the solution is called precipitation reaction.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

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

Oxidation - reduction reaction: The electrons are transferred to one to other is called oxidation reduction reaction.

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.

Soluble compounds in water

Almost all the salts of ${\text{Na}}^{\text{+}}{\text{, K}}^{\text{+}}{\text{, NH}}_{\text{4}}{}^{\text{+}}\text{,}\left({\text{NO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{4}}{}^{\text{-}}\right)\text{and}\left({\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right)$ are soluble.

Almost all the salts of ${\text{Cl}}^{\text{-}}{\text{, Br}}^{\text{-}}{\text{, I}}^{\text{-}}\left(\text{halides}\right)$ are soluble.  But some of the halides are $\left({\text{Ag}}^{\text{+}}{\text{, Hg}}_{\text{2}}{}^{\text{2+}}{\text{, Pb}}^{\text{2+}}\right)$ insoluble.

Salts of F^- are soluble.  But some of the fluoride salt of ${\text{Mg}}^{\text{2+}}{\text{, Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}$ are insoluble.

Salts of $\text{sulfate }\left({\text{SO}}_{\text{4}}{}^{\text{2-}}\right)$ are soluble.  But sulfates of ${\text{Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}{\text{, Ag}}^{\text{+}}\text{,}$ are insoluble.

Insoluble compounds in water:

Most of the salts of $\left({\text{CO}}_{\text{3}}{}^{\text{2-}}\right)\text{,}\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{, }\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)\text{,}\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)\text{and}\left({\text{S}}^{\text{2-}}\right)$ are insoluble but some of the salts of NH₄⁺, alkali metal cations and BaS are soluble.

Most of the metal hydroxides and oxides are insoluble in water bit some of the alkali metal hydroxides, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\text{and Sr}{\left(\text{OH}\right)}_{\text{2}}$ are soluble in water.

Answer to Problem 63PS

The given reaction is oxidation and reduction reaction and the state of reactant and product is shown below.

    

Explanation of Solution

The balancing of the reaction and the type of the reaction is shown below, the given reaction is oxidation and reduction reaction and the state of the reaction is shown below

    

The given compound is calcium and hydrogen bromide.  In this reaction hydrogen bromide reaction with calcium to give calcium bromide and hydrogen gas.  Here the oxidation state of calcium is zero in the reactant and $+2$ in the product, calcium loss the electron similarly the oxidation state of hydrogen is $+1$ in the reactant zero in the product.  Therefore it is oxidation-reduction reaction, and gas as hydrogen is evolved from the reaction.

     

Balance the equation,

Balance the hydrogen atom in the given 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 atoms in the left side.  Therefore two molecule of hydrogen bromide is added to left side of reaction.  Therefore the balanced equation is given below.

     

Interpretation Introduction

Interpretation:

• The given reaction has to be classified as precipitation or acid-base or gas forming reaction and the reaction has to be balanced.
• State of the product and reactant should be shown.

Concept introduction:

Precipitation reaction: The formation of the product is insoluble when the ions combine in the solution is called precipitation reaction.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

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

Oxidation - reduction reaction: The electrons are transferred to one to other is called oxidation reduction reaction.

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.

Soluble compounds in water

Almost all the salts of ${\text{Na}}^{\text{+}}{\text{, K}}^{\text{+}}{\text{, NH}}_{\text{4}}{}^{\text{+}}\text{,}\left({\text{NO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{3}}{}^{\text{-}}\right)\text{,}\left({\text{ClO}}_{\text{4}}{}^{\text{-}}\right)\text{and}\left({\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right)$ are soluble.

Almost all the salts of ${\text{Cl}}^{\text{-}}{\text{, Br}}^{\text{-}}{\text{, I}}^{\text{-}}\left(\text{halides}\right)$ are soluble.  But some of the halides are $\left({\text{Ag}}^{\text{+}}{\text{, Hg}}_{\text{2}}{}^{\text{2+}}{\text{, Pb}}^{\text{2+}}\right)$ insoluble.

Salts of F^- are soluble.  But some of the fluoride salt of ${\text{Mg}}^{\text{2+}}{\text{, Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}$ are insoluble.

Salts of $\text{sulfate }\left({\text{SO}}_{\text{4}}{}^{\text{2-}}\right)$ are soluble.  But sulfates of ${\text{Ca}}^{\text{2+}}{\text{, Sr}}^{\text{2+}}{\text{, Ba}}^{\text{2+}}{\text{, Pb}}^{\text{2+}}{\text{, Ag}}^{\text{+}}\text{,}$ are insoluble.

Insoluble compounds in water:

Most of the salts of $\left({\text{CO}}_{\text{3}}{}^{\text{2-}}\right)\text{,}\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{, }\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)\text{,}\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)\text{and}\left({\text{S}}^{\text{2-}}\right)$ are insoluble but some of the salts of NH₄⁺, alkali metal cations and BaS are soluble.

Most of the metal hydroxides and oxides are insoluble in water bit some of the alkali metal hydroxides, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\text{and Sr}{\left(\text{OH}\right)}_{\text{2}}$ are soluble in water.

Answer to Problem 63PS

The given reaction is precipitation reaction and the state of the reaction is shown below.

    

Explanation of Solution

The balancing of the reaction and the type of the reaction is shown below, the given reaction is precipitation reaction and the state of the reaction is shown below

    

The given compound is magnesium chloride and sodium hydroxide.  In this reaction magnesium chloride reaction with sodium hydroxide to give magnesium hydroxide and sodiumchloride.  Most of the metal hydroxides and oxides are insoluble in water bit some of the alkali metal hydroxides, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\text{and Sr}{\left(\text{OH}\right)}_{\text{2}}$ are soluble in water.  Therefore,

the given reaction is precipitation reaction.

    

Balance the equation,

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

    

Balance the oxygen atom in the given equation, there are two oxygen atoms in the right side and one oxygen atoms in the left side. Therefore two molecule of sodium hydroxide is added to left side of reaction. Therefore the balanced equation is given below.