Chapter 3, Problem 52QAP

Write a balanced equation for the reaction between

(a) dihydrogen sulfide and sulfur dioxide gases to form sulfur solid and steam.

(b) methane, ammonia, and oxygen gases to form hydrogen cyanide gas and steam.

(c) iron(lll) oxide and hydrogen gas to form molten iron and steam.

(d) uranium(IV) oxide and hydrogen fluoride gas to form uranium(IV) fluoride and steam.

(e) the combustion of ethyl alcohol (C₂H₅OH) to give carbon dioxide and water.

Interpretation Introduction

(a)

Interpretation:

The balanced chemical equation for the formation of sulfur solid and steam from dihydrogen sulfide and sulfur dioxide needs to be determined.

Concept introduction:

A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. A balanced chemical reaction is in accordance with the law of conservation of mass.

Answer to Problem 52QAP

${\text{2H}}_{\text{2}}\text{S}\left(g\right){\text{+SO}}_{\text{2}}\left(g\right)\to 3\text{S}\left(s\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The reaction of dihydrogen sulfide and sulfur dioxide gases to form sulfur solid and steam is represented as follows:

${\text{H}}_{\text{2}}\text{S}\left(g\right){\text{+SO}}_{\text{2}}\left(g\right)\to \text{S}\left(s\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)$

Give coefficient 2 to ${\text{H}}_{\text{2}}\text{O}$ to balance the number of oxygen atoms.

${\text{H}}_{\text{2}}\text{S}\left(g\right){\text{+SO}}_{\text{2}}\left(g\right)\to \text{S}\left(s\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Now, give coefficient 2 to ${\text{H}}_{\text{2}}\text{S}$ to balance the number of hydrogen atoms.

${\text{2H}}_{\text{2}}\text{S}\left(g\right){\text{+SO}}_{\text{2}}\left(g\right)\to \text{S}\left(s\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

In the last step, give coefficient 3 to S thus,

${\text{2H}}_{\text{2}}\text{S}\left(g\right){\text{+SO}}_{\text{2}}\left(g\right)\to 3\text{S}\left(s\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Therefore, the balanced chemical reaction will be:

${\text{2H}}_{\text{2}}\text{S}\left(g\right){\text{+SO}}_{\text{2}}\left(g\right)\to 3\text{S}\left(s\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced chemical equation for the formation of hydrogen cyanide gas and steam from methane, ammonia and oxygen needs to be determined.

Concept introduction:

A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. A balanced chemical reaction is in accordance with the law of conservation of mass.

Answer to Problem 52QAP

${\text{2CH}}_{\text{4}}\left(g\right){\text{+2NH}}_{\text{3}}\left(g\right){\text{+3O}}_{\text{2}}\left(g\right)\to 2\text{HCN}\left(g\right){\text{+6H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The reaction of methane, ammonia and oxygen gas to form hydrogen cyanide gas and steam is as follows:

${\text{CH}}_{\text{4}}\left(g\right){\text{+NH}}_{\text{3}}\left(g\right){\text{+O}}_{\text{2}}\left(g\right)\to \text{HCN}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)$

Give coefficient 3 to ${\text{H}}_{\text{2}}\text{O}$ to balance the number of hydrogen atoms

${\text{CH}}_{\text{4}}\left(g\right){\text{+NH}}_{\text{3}}\left(g\right){\text{+O}}_{\text{2}}\left(g\right)\to \text{HCN}\left(g\right){\text{+3H}}_{\text{2}}\text{O}\left(g\right)$

Now, give coefficient 3/2 to oxygen gas to balance the number of oxygen atoms.

${\text{CH}}_{\text{4}}\left(g\right){\text{+NH}}_{\text{3}}\left(g\right){\text{+3/2O}}_{\text{2}}\left(g\right)\to \text{HCN}\left(g\right){\text{+3H}}_{\text{2}}\text{O}\left(g\right)$

Or,

${\text{2CH}}_{\text{4}}\left(g\right){\text{+2NH}}_{\text{3}}\left(g\right){\text{+3O}}_{\text{2}}\left(g\right)\to 2\text{HCN}\left(g\right){\text{+6H}}_{\text{2}}\text{O}\left(g\right)$

Therefore, the balanced chemical reaction will be:

${\text{2CH}}_{\text{4}}\left(g\right){\text{+2NH}}_{\text{3}}\left(g\right){\text{+3O}}_{\text{2}}\left(g\right)\to 2\text{HCN}\left(g\right){\text{+6H}}_{\text{2}}\text{O}\left(g\right)$

Interpretation Introduction

(c)

Interpretation:

The balanced chemical equation for the formation of molten iron and steam from iron (III) oxide and hydrogen gas needs to be determined.

Concept introduction:

A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. A balanced chemical reaction is in accordance with the law of conservation of mass.

Answer to Problem 52QAP

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}\left(g\right)\to 2\text{Fe}\left(s\right){\text{+3H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The reaction of iron (III) oxide and hydrogen gas to form molten iron and steam is as follows:

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)\to \text{Fe}\left(s\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)$

Give coefficient 2 to Fe to balance the number of iron atoms.

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)\to 2\text{Fe}\left(s\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)$

Now, give coefficient 3 to ${\text{H}}_{\text{2}}\text{O}$ to balance the number of oxygen atoms.

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)\to 2\text{Fe}\left(s\right){\text{+3H}}_{\text{2}}\text{O}\left(g\right)$

In the last step, give coefficient 3 to hydrogen gas to balance the number of hydrogen atoms.

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}\left(g\right)\to 2\text{Fe}\left(s\right){\text{+3H}}_{\text{2}}\text{O}\left(g\right)$

Therefore, the balanced chemical reaction will be:

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}\left(g\right)\to 2\text{Fe}\left(s\right){\text{+3H}}_{\text{2}}\text{O}\left(g\right)$

Interpretation Introduction

(d)

Interpretation:

The balanced chemical equation for the formation of uranium (IV) fluoride and steam from uranium(IV) oxide and hydrogen fluoride needs to be determined.

Concept introduction:

A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. A balanced chemical reaction is in accordance with the law of conservation of mass.

Answer to Problem 52QAP

${\text{UO}}_{\text{2}}\left(g\right)\text{+4HF}\left(g\right)\to {\text{UF}}_{\text{4}}\left(g\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The reaction of uranium (IV) oxide with hydrogen fluoride gas to give uranium (IV) fluoride and steam is as follows:

${\text{UO}}_{\text{2}}\left(g\right)\text{+HF}\left(g\right)\to {\text{UF}}_{\text{4}}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)$

Give coefficient 4 to HF to balance the number of F atoms.

${\text{UO}}_{\text{2}}\left(g\right)\text{+4HF}\left(g\right)\to {\text{UF}}_{\text{4}}\left(g\right){\text{+H}}_{\text{2}}\text{O}\left(g\right)$

Now, give coefficient 2 to water to balance the number of hydrogen and oxygen atoms.

${\text{UO}}_{\text{2}}\left(g\right)\text{+4HF}\left(g\right)\to {\text{UF}}_{\text{4}}\left(g\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Therefore, the balanced chemical reaction will be:

${\text{UO}}_{\text{2}}\left(g\right)\text{+4HF}\left(g\right)\to {\text{UF}}_{\text{4}}\left(g\right){\text{+2H}}_{\text{2}}\text{O}\left(g\right)$

Interpretation Introduction

(e)

Interpretation:

The balanced chemical equation for the combustion of ethyl alcohol needs to be determined.

Concept introduction:

A chemical reaction is said to be balanced if all the constituent atoms are same on both sides of the reaction arrow. A balanced chemical reaction is in accordance with the law of conservation of mass.

Answer to Problem 52QAP

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right){\text{+3O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The combustion reaction of ethyl alcohol to give carbon dioxide and water is as follows:

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right){\text{+O}}_{\text{2}}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Give coefficient 2 to carbon dioxide to balance the number of C atoms.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right){\text{+O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Now, give coefficient 3 to water to balance the number of hydrogen atoms.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right){\text{+O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

In the last step, give coefficient 3 to oxygen gas to balance the number of oxygen atoms.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right){\text{+3O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Therefore, the balanced chemical reaction will be:

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right){\text{+3O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$