Chapter 3, Problem 3.35P

(a)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced and name of reactants, and products has to be given.

    $\text{Al}(s)+{\text{Cl}}_{\text{2}}(g)\to {\text{AlCl}}_{\text{3}}(s)$

Concept Introduction:

Chemical reaction is represented in short form by using the molecular formulas of products and reactants in which an arrow is present between reactants and products is known as chemical equation.  Atoms are neither created nor destroyed in any chemical reaction.  Therefore, the resulting chemical equation written for the chemical reaction should also be balanced.

Steps followed to obtain balanced chemical equation:

• Chemical formulas of the products and reactants are written.  Individual reactants and products are separated using plus sign while the reactants are separated from products by an arrow.
• Elements that appear only once on the reactant side and product side is balanced first.  Subscript should not be changed in chemical formula while balancing coefficients can be added until the total number of atoms of an element becomes equal on reactant and product side.
• Other elements present also balanced in the same way by adding balancing coefficients.
• Final check has to be performed in order to check that each element that is present in the chemical equation is balanced.
• Symbols have to be added for solids, liquids gases and aqueous solutions to indicate the physical state of the known reactants and products.

Explanation of Solution

Chemical equation given is shown below.

    $\text{Al}(s)+{\text{Cl}}_{\text{2}}(g)\to {\text{AlCl}}_{\text{3}}(s)$

Above chemical equation is not balanced.  There are two chlorine atoms on the left side of the equation while there are three chlorine atoms on right side of equation.  Adding coefficient $3$ before ${\text{Cl}}_{\text{2}}$ in the reactant side and $2$ before ${\text{AlCl}}_{\text{3}}$ in the product side balances the chlorine atom.  Obtained equation after adding coefficient is given as shown below.

    $\text{Al}(s)+3{\text{Cl}}_{\text{2}}(g)\to 2{\text{AlCl}}_{\text{3}}(s)(notbalanced)$

In the above equation, there is one aluminium atom in reactant side while there are two aluminium atoms on the product side.  Adding coefficient $2$ before $\text{Al}$ in the reactant side balances the aluminium atom on both sides of the equation.  The balanced chemical equation is given as shown below.

    $\text{2Al}(s)+3{\text{Cl}}_{\text{2}}(g)\to 2{\text{AlCl}}_{\text{3}}(s)(balanced)$

Name of reactants and products is given below.

    $\begin{array}{l}\text{Al}\text{-}\text{Aluminium}\\ {\text{Cl}}_{\text{2}}\text{-}\text{Chlorine}\\ {\text{AlCl}}_{\text{3}}\text{-}\text{Aluminium}\text{chloride}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced and name of reactants, and products has to be given.

    $\text{Al}(s)+{\text{O}}_{\text{2}}(g)\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

    $\text{Al}(s)+{\text{O}}_{\text{2}}(g)\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)$

Above chemical equation is not balanced.  There are two oxygen atoms on the left side of the equation while there are three oxygen atoms on right side of equation.  Adding coefficient $3$ before ${\text{O}}_{\text{2}}$ in the reactant side and $2$ before ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ in the product side balances the oxygen atom.  Obtained equation after adding coefficient is given as shown below.

    $\text{Al}(s)+3{\text{O}}_{\text{2}}(g)\to 2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)(notbalanced)$

In the above equation, there is one aluminium atom in reactant side while there are four aluminium atoms on the product side.  Adding coefficient $4$ before $\text{Al}$ in the reactant side balances the aluminium atom on both sides of the equation.  The balanced chemical equation is given as shown below.

    $\text{4Al}(s)+3{\text{O}}_{\text{2}}(g)\to 2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}(s)(balanced)$

Name of reactants and products is given below.

    $\begin{array}{l}\text{Al}\text{-}\text{Aluminium}\\ {\text{O}}_{\text{2}}\text{-}\text{Oxygen}\\ {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{-}\text{Aluminium}\text{oxide}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced and name of reactants, and products has to be given.

    ${\text{COCl}}_{\text{2}}(g)+\text{Na}(s)\to \text{NaCl}(s)+\text{CO}(g)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

    ${\text{COCl}}_{\text{2}}(g)+\text{Na}(s)\to \text{NaCl}(s)+\text{CO}(g)$

Above chemical equation is not balanced.  There are two chlorine atoms on the left side of the equation while there is only one chlorine atom on the right side of the equation.  Adding coefficient $2$ before $\text{NaCl}$ on the product side balances the chlorine atom on both sides of equation.  The chemical equation obtained is shown as given below.

    ${\text{COCl}}_{\text{2}}(g)+\text{Na}(s)\to 2\text{NaCl}(s)+\text{CO}(g)(notbalanced)$

In the above chemical equation, there is one sodium atom on the left side of the equation while two sodium atoms are present on right side of the equation.  Adding coefficient $2$ before $\text{Na}$ in the reactant side balances the sodium atoms on both sides of the equation.  The balanced chemical equation obtained is shown as given below.

    ${\text{COCl}}_{\text{2}}(g)+2\text{Na}(s)\to 2\text{NaCl}(s)+\text{CO}(g)(balanced)$

Name of reactants and products is given below.

    $\begin{array}{l}{\text{COCl}}_{\text{2}}\text{-}\text{Phosgene}\\ \text{Na}\text{-}\text{Sodium}\\ \text{NaCl}\text{-}\text{Sodium}\text{chloride}\\ \text{CO}\text{-}\text{Carbon monoxide}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced and name of reactants, and products has to be given.

    $\text{Be}(s)+{\text{O}}_{\text{2}}(g)\to \text{BeO}(s)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

    $\text{Be}(s)+{\text{O}}_{\text{2}}(g)\to \text{BeO}(s)$

Above chemical equation is not balanced.  There are two oxygen atoms on the left side of the equation while there is only one oxygen atom on right side of equation.  Adding coefficient $2$ before $\text{BeO}$ in the product side balances the oxygen atom.  Obtained equation after adding coefficient is given as shown below.

    $\text{Be}(s)+{\text{O}}_{\text{2}}(g)\to 2\text{BeO}(s)(notbalanced)$

In the above equation, there is one beryllium atom in reactant side while there are two beryllium atoms on the product side.  Adding coefficient $2$ before $\text{Be}$ in the reactant side balances the beryllium atom on both sides of the equation.  The balanced chemical equation is given as shown below.

    $\text{2Be}(s)+{\text{O}}_{\text{2}}(g)\to 2\text{BeO}(s)(balanced)$

Name of reactants and products is given below.

    $\begin{array}{l}\text{Be}\text{-}\text{Beryllium}\\ {\text{O}}_{\text{2}}\text{-}\text{Oxygen}\\ \text{BeO}\text{-}\text{Beryllium oxide}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Chemical equation given below has to be balanced and name of reactants, and products has to be given.

    $\text{K}(s)+\text{S}(l)\to {\text{K}}_{\text{2}}\text{S}(s)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Chemical equation given is shown below.

    $\text{K}(s)+\text{S}(l)\to {\text{K}}_{\text{2}}\text{S}(s)$

Above chemical equation is not balanced.  There is one potassium atom on the reactant side while there are two potassium atoms on the product side.  Adding coefficient $2$ before $\text{K}$ in the reactant side balances the potassium atom on both sides of the equation.  The balanced chemical equation is given as shown below.

    $\text{2K}(s)+\text{S}(l)\to {\text{K}}_{\text{2}}\text{S}(s)(balanced)$

Name of reactants and products is given below.

    $\begin{array}{l}\text{K}\text{-}\text{Potassium}\\ \text{S}\text{-}\text{Sulfur}\\ {\text{K}}_{\text{2}}\text{S}\text{-}\text{Potassium sulfide}\end{array}$