Chapter 2, Problem 2.35QP

(a)

Interpretation Introduction

Interpretation:

The correct element and/or compound symbols, formulas, and coefficients needed to produce complete, balanced equation has to be given.

Answer to Problem 2.35QP

$\text{2Li}\text{+}{\text{Cl}}_{\text{2}}\stackrel{}{\to }\text{2LiCl}$

Explanation of Solution

The product in the given chemical equation is Lithium chloride.  Hence, the starting material has to be Lithium and Chlorine.

$\text{Li}\text{+}{\text{Cl}}_{\text{2}}\stackrel{}{\to }\text{LiCl}$

To balance the chemical equation, coefficient 2 has to be added before $\text{Li}$ and $\text{LiCl}$.

$\text{2Li}\text{+}{\text{Cl}}_{\text{2}}\stackrel{}{\to }\text{2LiCl}$

(b)

Interpretation Introduction

Interpretation:

The correct element and/or compound symbols, formulas, and coefficients needed to produce complete, balanced equation has to be given.

Answer to Problem 2.35QP

$\text{16Na}\text{+}{\text{S}}_{\text{8}}\stackrel{}{\to }{\text{8Na}}_{\text{2}}\text{S}$

Explanation of Solution

The product in the given chemical equation is sodium sulfide.  Hence, the starting material has to be Sodium and Sulfur.

$\text{Na}\text{+}\text{S}\stackrel{}{\to }{\text{Na}}_{\text{2}}\text{S}$

To balance the chemical equation, coefficient 16 before sodium, and coefficient 8 has to be added before ${\text{Na}}_{\text{2}}\text{S}$.  For the sulfur atom in the reactant side, the solid sulfur can be considered.  Hence, sulfur is entered as ${\text{S}}_{\text{8}}$.

$\text{16Na}\text{+}{\text{S}}_{\text{8}}\stackrel{}{\to }{\text{8Na}}_{\text{2}}\text{S}$

(c)

Interpretation Introduction

Interpretation:

The correct element and/or compound symbols, formulas, and coefficients needed to produce complete, balanced equation has to be given.

Answer to Problem 2.35QP

${\text{3Al}}_{\text{2}}\text{+}{\text{3I}}_{\text{2}}\stackrel{}{\to }2{\text{AlI}}_{\text{3}}$

Explanation of Solution

The product in the given chemical equation is aluminium iodide.  Hence, the starting material has to be Aluminium and Iodine.

${\text{Al}}_{\text{2}}\text{+}{\text{I}}_{\text{2}}\stackrel{}{\to }{\text{AlI}}_{\text{3}}$

To balance the chemical equation, coefficient 3 before aluminium and iodine.  Coefficient 2 has to be added before ${\text{AlI}}_{\text{3}}$.  Hence, the equation can be written as,

${\text{3Al}}_{\text{2}}\text{+}{\text{3I}}_{\text{2}}\stackrel{}{\to }2{\text{AlI}}_{\text{3}}$

(d)

Interpretation Introduction

Interpretation:

The correct element and/or compound symbols, formulas, and coefficients needed to produce complete, balanced equation has to be given.

Answer to Problem 2.35QP

$\text{3Ba}\text{+}{\text{N}}_{\text{2}}\stackrel{}{\to }{\text{Ba}}_{\text{3}}{\text{N}}_{\text{2}}$

Explanation of Solution

The product in the given chemical equation is Barium nitride.  Hence, the starting material has to be Barium and Nitrogen.

$\text{Ba}\text{+}{\text{N}}_{\text{2}}\stackrel{}{\to }{\text{Ba}}_{\text{3}}{\text{N}}_{\text{2}}$

To balance the chemical equation, coefficient 3 has to be added before barium.  Hence, the equation can be written as,

$\text{3Ba}\text{+}{\text{N}}_{\text{2}}\stackrel{}{\to }{\text{Ba}}_{\text{3}}{\text{N}}_{\text{2}}$

(e)

Interpretation Introduction

Interpretation:

The correct element and/or compound symbols, formulas, and coefficients needed to produce complete, balanced equation has to be given.

Answer to Problem 2.35QP

$\text{12V}\text{+}{\text{5P}}_{\text{4}}\stackrel{}{\to }{\text{4V}}_{\text{3}}{\text{P}}_{\text{5}}$

Explanation of Solution

The product in the given chemical equation is ${\text{V}}_{\text{3}}{\text{P}}_{\text{5}}$.  Hence, the starting material has to be Vanadium and Phosphorous.

$\text{V}\text{+}{\text{P}}_{\text{4}}\stackrel{}{\to }{\text{V}}_{\text{3}}{\text{P}}_{\text{5}}$

To balance the chemical equation, start balancing phosphorous first.  In the product side coefficient 4 has to be added and in the reactant side, coefficient 5 has to be added before phosphorous.  Hence, the equation can be written as,

$\text{V}\text{+}5{\text{P}}_{\text{4}}\stackrel{}{\to }4{\text{V}}_{\text{3}}{\text{P}}_{\text{5}}$

Now start to balance vanadium.  This is done by adding coefficient 12 before vanadium in the reactant side.  The balanced equation can be written as,

$\text{12V}\text{+}{\text{5P}}_{\text{4}}\stackrel{}{\to }{\text{4V}}_{\text{3}}{\text{P}}_{\text{5}}$