Chapter 4, Problem 63E

For each chemical reaction, draw in the missing molecules necessary to balance their equations. Carbon atoms are black, hydrogen atoms are white, oxygen atoms are red, and nitrogen atoms are blue.

Interpretation Introduction

Interpretation:

The chemical reactions are to be balanced for the given molecular model reactions, where carbon atoms are black, hydrogen atoms are white, oxygen atoms are red, and nitrogen atoms are blue.

Concept Introduction:

A chemical equation is considered to be the symbolic depiction of a chemical reaction, which 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 the molecule, but not their structural arrangement

The molecular model represents the molecules in a three-dimensional configuration.

Answer to Problem 63E

Solution: The chemical formulas are represented as follows: a) ${\text{C}}_2{\text{H}}_{\text{4}}+3{\text{O}}_2\to 2{\text{CO}}_2+2{\text{H}}_2\text{O}$ b) ${\text{N}}_2+3{\text{H}}_2\to 2{\text{NH}}_3$ c) $2\text{NO}+2\text{CO}\to {\text{N}}_{\text{2}}+2{\text{CO}}_2$ d) $4{\text{NO}}_2+{\text{O}}_2+2{\text{H}}_2\text{O}\to 4{\text{HNO}}_3$

Explanation of Solution

a)

In the given molecular model, the reaction can be represented in a chemical reaction as follows:

${\text{C}}_2{\text{H}}_{\text{4}}+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$

$\text{C}$ atom is balanced by multiplying with a coefficient $2$ in ${\text{CO}}_2$ as follows:

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

In the next step, the $\text{O}$ atom is balanced by the multiplying with a coefficient $3$ in ${\text{O}}_2$ as follows:

${\text{C}}_2{\text{H}}_{\text{4}}+3{\text{O}}_2\to 2{\text{CO}}_2+{\text{H}}_2\text{O}$

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

${\text{C}}_2{\text{H}}_{\text{4}}+3{\text{O}}_2\to 2{\text{CO}}_2+2{\text{H}}_2\text{O}$

b)

In the given molecular model, the reaction can be represented in a chemical reaction as follows:

${\text{N}}_2+{\text{H}}_2\to {\text{NH}}_3$

$\text{N}$ atom is balanced by multiplying with a coefficient $2$ in ${\text{NH}}_{\text{3}}$ as follows:

${\text{N}}_2+{\text{H}}_2\to 2{\text{NH}}_3$

In the next step, the $\text{H}$ atom is balanced by the multiplying with a coefficient $3$ in ${\text{H}}_{\text{2}}$ as follows:

${\text{N}}_2+3{\text{H}}_2\to 2{\text{NH}}_3$

c)

In the given molecular model, the reaction can be represented in a chemical reaction as follows:

$\text{NO}+\text{CO}\to {\text{N}}_{\text{2}}+{\text{CO}}_{\text{2}}$

$\text{N}$ atom is balanced by multiplying with a coefficient $2$ in $\text{NO}$ as follows:

$2\text{NO}+\text{CO}\to {\text{N}}_{\text{2}}+{\text{CO}}_2$

In next step, the $\text{O}$ atom is balanced by multiplying with a coefficient $2$ in ${\text{CO}}_2$ of product part as follows:

$2\text{NO}+\text{CO}\to {\text{N}}_{\text{2}}+2{\text{CO}}_2$

Now, in the final step, the $\text{C}$ atom is balanced by multiplying with a coefficient $2$ in CO of reactant part as follows:

$2\text{NO}+2\text{CO}\to {\text{N}}_{\text{2}}+2{\text{CO}}_2$

d)

In the given molecular model, the reaction can be represented in a chemical reaction as follows:

${\text{NO}}_2+{\text{O}}_2{\text{+H}}_2\text{O}\to {\text{HNO}}_3$

$\text{H}$ atom is balanced by multiplying with a coefficient $4$ in ${\text{HNO}}_{\text{3}}$ of product part while coefficient $2$ in ${\text{H}}_{\text{2}}\text{O}$ as follows:

${\text{NO}}_2+{\text{O}}_2+2{\text{H}}_2\text{O}\to +4{\text{HNO}}_3$

In the next step, $\text{O}$ and $\text{N}$ are balanced by multiplying with a coefficient $4$ in ${\text{NO}}_2$ as follows:

$4{\text{NO}}_2+{\text{O}}_2+2{\text{H}}_2\text{O}\to 4{\text{HNO}}_3$

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

$4{\text{NO}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}+4{\text{HNO}}_3$