Chapter 4, Problem 52E

Interpretation Introduction

Interpretation:

The chemical equations are to be balanced for the given reactions.

Concept introduction:

A chemical equation is a symbolic depiction of a chemical reaction that contains symbols and formula.

In a chemical reaction, the starting substances on the left side are known as reactants, while the substances on the right side are known as products.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 52E

Solution: The balanced chemical equation is given as follows:

$2{\text{C}}_2{\text{H}}_6+7{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$

$\text{2}{\text{H}}_{\text{2}}{\text{S+SO}}_{\text{2}}\to \text{3}\text{S+2}{\text{H}}_{\text{2}}\text{O}$

$2{\text{CH}}_3\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

a) ${\text{C}}_2{\text{H}}_6+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$

$\text{H}$ atom is balanced by multiplying the coefficient $6$ with ${\text{H}}_2\text{O}$ and the coefficient $2$ with ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ as follows:

$2{\text{C}}_2{\text{H}}_6+{\text{O}}_2\to {\text{CO}}_2+6{\text{H}}_2\text{O}$

In the next step, $\text{C}$ atom is balanced by multiplying the coefficient $4$ with ${\text{CO}}_2$ as follows:

$2{\text{C}}_2{\text{H}}_6+{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$

Now, in the final step, $\text{O}$ atom is balanced by multiplying the coefficient $7$ with ${\text{O}}_2$ in the reactant part as follows:

$2{\text{C}}_2{\text{H}}_6+7{\text{O}}_2\to 4{\text{CO}}_2+6{\text{H}}_2\text{O}$

b) ${\text{H}}_{\text{2}}{\text{S+SO}}_{\text{2}}\to {\text{S+H}}_{\text{2}}\text{O}$

$\text{O}$ atom is balanced by multiplying the coefficient $2$ with ${\text{H}}_2\text{O}$, which is present in the product part as follows:

${\text{H}}_{\text{2}}{\text{S+SO}}_{\text{2}}\to \text{S+2}{\text{H}}_{\text{2}}\text{O}$

In the next step, $\text{H}$ atom is balanced by multiplying the coefficient $2$ with ${\text{H}}_{\text{2}}\text{S}$ in the reactant part as follows:

$\text{2}{\text{H}}_{\text{2}}{\text{S+SO}}_{\text{2}}\to \text{S+2}{\text{H}}_{\text{2}}\text{O}$

Now, in the final step, $\text{S}$ atom is balanced by multiplying the coefficient $3$ with $\text{S}$ in the product part as follows:

$\text{2}{\text{H}}_{\text{2}}{\text{S+SO}}_{\text{2}}\to \text{3}\text{S+2}{\text{H}}_{\text{2}}\text{O}$

c) ${\text{CH}}_3\text{OH}+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

$\text{H}$ atom is balanced by multiplying the coefficient $4$ with ${\text{H}}_2\text{O}$ and the coefficient $2$ with ${\text{CH}}_{\text{3}}\text{OH}$ as follows:

$2{\text{CH}}_3\text{OH}+{\text{O}}_2\to {\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$

In the next step, $\text{C}$ atom is balanced by multiplying the coefficient $2$ with ${\text{CO}}_2$ as follows:

$2{\text{CH}}_3\text{OH}+{\text{O}}_2\to 2{\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$

Now, in the final step, $\text{O}$ atom is balanced by multiplying the coefficient $3$ with ${\text{O}}_{\text{2}}$ in the reactant part as follows:

$2{\text{CH}}_3\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$