Chapter 11.3, Problem 27SSC

(a)

Interpretation Introduction

Interpretation:

Whether in the given reaction 4 moles of ${\text{P}}_{\text{4}}$ reacts with 1.5 moles of ${\text{S}}_8$ to form 4 moles of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ or not needs to be determined.

${\text{8 P}}_{\text{4}}{\text{ + 3 S}}_8\stackrel{}{\to }{\text{8 P}}_{\text{4}}{\text{S}}_{\text{3}}$

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactants, which is present in limited amount and determines the amount of product is known as limiting reactant. Whereas another reactant is known as excess reactant.

Answer to Problem 27SSC

4 moles of ${\text{P}}_{\text{4}}$ reacts with 1.5 moles of ${\text{S}}_8$ to form 4 moles of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ is a correct statement.

Explanation of Solution

${\text{8 P}}_{\text{4}}{\text{ + 3 S}}_8\stackrel{}{\to }{\text{8 P}}_{\text{4}}{\text{S}}_{\text{3}}$

Moles of ${\text{P}}_{\text{4}}$ = 4 moles

Moles of ${\text{S}}_8$ = 1.5 moles

Moles of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ = 4 moles

From the balance chemical reaction:

$\begin{array}{l}{\text{4 moles P}}_{\text{4}}\text{×}\frac{{\text{3 moles S}}_{\text{8}}}{{\text{8 moles P}}_{\text{4}}}\text{=1}{\text{.5 moles S}}_{\text{8}}\\ \text{1}{\text{.5 moles S}}_{\text{8}}\text{×}\frac{{\text{8 moles P}}_{\text{4}}{\text{S}}_{\text{3}}}{{\text{3 moles S}}_{\text{8}}}{\text{=4 moles P}}_{\text{4}}{\text{S}}_{\text{3}}\end{array}$

Thus 4 moles of ${\text{P}}_{\text{4}}$ reacts with 1.5 moles of ${\text{S}}_8$ to form 4 moles of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ is a correct statement.

(b)

Interpretation Introduction

Interpretation:

Whether sulfur is a limiting reactant or not needs to be determined.

${\text{8 P}}_{\text{4}}{\text{ + 3 S}}_8\stackrel{}{\to }{\text{8 P}}_{\text{4}}{\text{S}}_{\text{3}}$

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactants, which is present in limited amount and determines the amount of product is known as limiting reactant. Whereas another reactant is known as excess reactant.

Answer to Problem 27SSC

“When 4 moles of ${\text{P}}_{\text{4}}$ reacts with 4 moles of ${\text{S}}_8$, ${\text{P}}_{\text{4}}$ is a limiting reactant.”

Explanation of Solution

${\text{8 P}}_{\text{4}}{\text{ + 3 S}}_8\stackrel{}{\to }{\text{8 P}}_{\text{4}}{\text{S}}_{\text{3}}$

Moles of ${\text{P}}_{\text{4}}$ = 4 moles

Moles of ${\text{S}}_8$ = 1.5 moles

Moles of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ = 4 moles

From the balance chemical reaction:

${\text{4 moles P}}_{\text{4}}\text{×}\frac{{\text{3 moles S}}_{\text{8}}}{{\text{8 moles P}}_{\text{4}}}\text{=1}{\text{.5 moles S}}_{\text{8}}$

Thus 4 moles of ${\text{P}}_{\text{4}}$ reacts with 1.5 moles of ${\text{S}}_8$ and with 4 moles of ${\text{S}}_8$, it will be excess reactant and ${\text{P}}_{\text{4}}$ must be limiting reactant. Thus the statement should be:

“When 4 moles of ${\text{P}}_{\text{4}}$ reacts with 4 moles of ${\text{S}}_8$, ${\text{P}}_{\text{4}}$ is a limiting reactant.”

(c)

Interpretation Introduction

Interpretation:

Whether 1320 g of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ forms or not if 6 moles of ${\text{P}}_{\text{4}}$ reacts with 6 moles of ${\text{S}}_8$ in the given below reaction needs to be determined.

${\text{8 P}}_{\text{4}}{\text{ + 3 S}}_8\stackrel{}{\to }{\text{8 P}}_{\text{4}}{\text{S}}_{\text{3}}$

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactants, which is present in limited amount and determines the amount of product is known as limiting reactant. Whereas another reactant is known as excess reactant.

Answer to Problem 27SSC

6 moles of ${\text{P}}_{\text{4}}$ reacts with 6 moles of ${\text{S}}_8$ in the given below reaction, 1320 g of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ forms.

Explanation of Solution

${\text{8 P}}_{\text{4}}{\text{ + 3 S}}_8\stackrel{}{\to }{\text{8 P}}_{\text{4}}{\text{S}}_{\text{3}}$

Moles of ${\text{P}}_{\text{4}}$ = 6 moles

Moles of ${\text{S}}_8$ = 8 moles

Mass of ${\text{P}}_{\text{4}}{\text{S}}_{\text{3}}$ = 1320 g

From the balance chemical reaction:

${\text{6 moles P}}_{\text{4}}\text{×}\frac{{\text{3 moles S}}_{\text{8}}}{{\text{8 moles P}}_{\text{4}}}\text{=2}{\text{.25 moles S}}_{\text{8}}$

${\text{P}}_{\text{4}}$ is limiting reactant and will determine the amount of product form.

${\text{6 moles P}}_{\text{4}}\text{×}\frac{{\text{8 moles P}}_{\text{4}}{\text{S}}_{\text{3}}}{{\text{8 moles P}}_{\text{4}}}\text{×}\frac{{\text{220 g P}}_{\text{4}}{\text{S}}_{\text{3}}}{{\text{1 moles P}}_{\text{4}}{\text{S}}_{\text{3}}}{\text{= 1320 g P}}_{\text{4}}{\text{S}}_{\text{3}}$

Thus given statement is correct.