Chapter 3, Problem 24QP

Interpretation Introduction

Interpretation:

The given chemical equations need to be balanced using the method mentioned in section 3.3.

Concept introduction:

Chemical equations are symbolic representations of chemical reactions.

On the left side of arrow, the reactants are to be written and on the right side ofarrow, the products are to be written.

Chemical equations are denoted using chemical formulae of the elements and compounds involved.

Chemical formulae of elements are written on the basis of their atomicity.

Atomicity is measured as the number of atoms present in the molecules of an element.

Chemical formulae of compounds are written on the basis of the molecularity.

Molecularity is the number of molecules that react in an elementary (single-step) reaction.

When the same numbers of atoms of every element on both sides of the arrow are present, the equation is said to be balanced.

Balancing of chemical equations is done on the principle of “The Law of Conservation of Mass.”

Answer to Problem 24QP

Solution:

a) ${\text{2C + O}}_{\text{2}}\to \text{ 2CO}$

b) ${\text{2CO + O}}_{\text{2}}\to {\text{ 2CO}}_{\text{2}}$

c) ${\text{H}}_{\text{2}}{\text{ + Br}}_{\text{2}}\to \text{ 2HBr}$

d) ${\text{2K + 2H}}_{\text{2}}\text{O }\to {\text{ 2KOH + H}}_{\text{2}}$

e) ${\text{2Mg + O}}_{\text{2}}\to \text{ 2MgO}$

f) ${\text{2O}}_3\to {\text{ 3O}}_{\text{2}}$

g) $\text{2H}\text{O}{}_2\to {\text{ 2H}}_{\text{2}}{\text{O + O}}_{\text{2}}$

h) $\text{N}{\text{+ 3H}}_{\text{2}}\to {\text{ 2NH}}_3$

i) $\text{Zn + 2AgCl }\to {\text{ ZnCl}}_{\text{2}}\text{ + 2Ag}$

j) ${\text{S}}_{\text{8}}{\text{ + 8O}}_{\text{2}}\to {\text{ 8SO}}_{\text{2}}$

k) ${\text{2NaOH + H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{ Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$

l) ${\text{Cl}}_{\text{2}}\text{ + 2NaI }\to {\text{ 2NaCl + I}}_{\text{2}}$

m) ${\text{3KOH + H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{ K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{ + 3H}}_{\text{2}}\text{O}$

n) ${\text{CH}}_{\text{4}}{\text{ + 4Br}}_{\text{2}}\to {\text{ CBr}}_{\text{4}}\text{ + 4HBr}$

Explanation of Solution

a) ${\text{C + O}}_{\text{2}}\to \text{ CO}$

There is one atom of carbon each on the right and the left side of the equation. However, on the left, there are two oxygen atoms, while on the right, there is only one oxygen atom. The equation is balanced by adding the coefficient $\text{2}$ in front of $\text{C}$ and $\text{CO}$.

The balanced equation is written as follows:

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

b) ${\text{CO + O}}_{\text{2}}\to {\text{ CO}}_{\text{2}}$

There is one atom of carbon each on the right and the left side of the equation. However, on the left, there are three oxygen atoms, while on the right, there aretwo oxygen atoms. The equation is balanced by adding the coefficient $\text{2}$ in front of $\text{CO}$ and ${\text{CO}}_{\text{2}}$.

The balanced equation is written as follows:

${\text{2CO + O}}_{\text{2}}\to {\text{ 2CO}}_{\text{2}}$

c) ${\text{H}}_{\text{2}}{\text{ + Br}}_{\text{2}}\to \text{ HBr}$

There are two atoms of hydrogen and two atoms of bromine on the left side of the equation. However, there is only atom in each of the two elements on the right side of the equation. The entire equation is balanced by adding the coefficient $\text{2}$ in front of $\text{HBr}$.

The balanced equation is written as follows:

${\text{H}}_{\text{2}}{\text{ + Br}}_{\text{2}}\to \text{ 2HBr}$

d) ${\text{K + H}}_{\text{2}}\text{O }\to {\text{ KOH + H}}_{\text{2}}$

There is one atom of potassium each on the right and the left side of the equation. There aretwo atoms of hydrogenon the left hand side but there are three on the right hand side. The number of oxygen atoms present on either side of the equation is one. The equation is balanced by adding the coefficient $\text{2}$ in front of $\text{K}$, ${\text{H}}_{\text{2}}\text{O}$ and $\text{KOH}$.

The balanced equation is written as follows:

${\text{2K + 2H}}_{\text{2}}\text{O }\to {\text{ 2KOH + H}}_{\text{2}}$

e) ${\text{Mg + O}}_{\text{2}}\to \text{ MgO}$

There is one atom of magnesium each on the left hand side and the right hand side of the equation. There are two atoms of oxygen on the left hand side but one on the right hand side. The equation is balanced by adding the coefficient $\text{2}$ in front of both $\text{Mg}$ and $\text{MgO}$.

The balanced equation is written as follows:

${\text{2Mg + O}}_{\text{2}}\to \text{ 2MgO}$

f) ${\text{O}}_3\to {\text{ O}}_{\text{2}}$

There are three atoms of oxygen on the left hand side but two on the right hand side of the equation. The equation is balanced by adding the coefficient $\text{2}$ in front of both ${\text{O}}_{\text{3}}$ and $\text{O}$.

The balanced equation is written as follows:

${\text{2O}}_3\to {\text{ 3O}}_{\text{2}}$

g) $\text{H}\text{O}{}_2\to {\text{ H}}_{\text{2}}{\text{O + O}}_{\text{2}}$

There are two atoms of hydrogen both on the right side and the left side of the equation. However, there are two atoms of oxygen on the left side and three on the right side of the equation. The equation is balanced by adding the coefficient $\text{2}$ in front of both ${\text{H}}_{\text{2}}{\text{O}}_2$ and ${\text{H}}_{\text{2}}\text{O}$.

The balanced equation is written as follows:

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

h) $\text{N}{\text{+ H}}_{\text{2}}\to {\text{ NH}}_3$

There are two atoms of nitrogen on the left hand side but one on the right hand side of the equation. Also, there are two atoms of hydrogen on the left side but three on the right side of the equation. The equation is balanced by adding the coefficient $3$ in front of ${\text{H}}_{\text{2}}$ and $2$ in front of ${\text{NH}}_{\text{3}}$.

The balanced equation is written as follows:

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

i) $\text{Zn + AgCl }\to {\text{ ZnCl}}_{\text{2}}\text{ + Ag}$

One atom of zinc is present both on the left side and the right side of the equation. Also, there is one atom of silver on both sides. However, on the left side, there is one atom of chlorine but on the right side, there are two. The equation is balanced by adding the coefficient $2$ both in front of $\text{AgCl}$ and $\text{Ag}$.

The balanced equation is written as follows:

$\text{Zn + 2AgCl }\to {\text{ ZnCl}}_{\text{2}}\text{ + 2Ag}$

j) ${\text{S}}_{\text{8}}{\text{ + O}}_{\text{2}}\to {\text{ SO}}_{\text{2}}$

There are eight atoms of sulfur on the left hand side but one on the right hand side of the equation. However, the number of oxygen atoms is two on both sides. The equation is balanced by adding the coefficient $8$ both in front of ${\text{O}}_{\text{2}}$ and ${\text{SO}}_{\text{2}}$.

The balanced equation is written as follows:

${\text{S}}_{\text{8}}{\text{ + 8O}}_{\text{2}}\to {\text{ 8SO}}_{\text{2}}$

k) ${\text{NaOH + H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{ Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + H}}_{\text{2}}\text{O}$

There is one atom of sodium on the left side while there are two atoms of sodium on the right side of the equation. The number of hydrogen atoms on the left side is three while that on the right side is two. There is one atom of sulfur and five atoms of oxygen on both sides. The equation is balanced by adding the coefficient $2$ both in front of $\text{NaOH}$ and ${\text{H}}_{\text{2}}\text{O}$. The balanced equation is written as follows:

${\text{2NaOH + H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{ Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$

l) ${\text{Cl}}_{\text{2}}\text{ + NaI }\to {\text{ NaCl + I}}_{\text{2}}$

There are two atoms of chlorine on the left, while there is one atom of chlorine on the right hand side of the equation. The number of sodium atoms is one on both sides. The number of iodine atoms on the left and the right side is $1$ and $2$, respectively. The equation is balanced by adding the coefficient $2$ both in front of $\text{NaI}$ and $\text{NaCl}$.

The balanced equation is written as follows:

${\text{Cl}}_{\text{2}}\text{ + 2NaI }\to {\text{ 2NaCl + I}}_{\text{2}}$

m) ${\text{KOH + H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{ K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{ + H}}_{\text{2}}\text{O}$

There is one atom of potassium on the left side and three atoms of potassium on the right side of the equation. The number of hydrogen atoms on the left and the right sidesarefour and two respectively. Phosphorus atom is one in number on both the sides. Also, oxygen atoms are five in number on both the sides. The equation is balanced by adding the coefficient $3$ both in front of $\text{KOH}$ and ${\text{H}}_{\text{2}}\text{O}$.

The balanced equation is written as follows:

${\text{3KOH + H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{ K}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{ + 3H}}_{\text{2}}\text{O}$

n) ${\text{CH}}_{\text{4}}{\text{ + Br}}_{\text{2}}\to {\text{ CBr}}_{\text{4}}\text{ + HBr}$

There is one atom of carbon both on the left and the right side of the equation. However, there are four atoms of hydrogen and two atoms of bromine on the left side, and one atom of hydrogen and five atoms of bromine on the right side of the equation. The equation is balanced by adding the coefficient $4$ both in front of ${\text{Br}}_{\text{2}}$ and $\text{HBr}$.

The balanced equation is written as follows:

${\text{CH}}_{\text{4}}{\text{ + 4Br}}_{\text{2}}\to {\text{ CBr}}_{\text{4}}\text{ + 4HBr}$