Chapter 3, Problem 25QP

Interpretation Introduction

Interpretation:

The given chemical equations are need to be balanced.

Concept Introduction:

Chemical equations are symbolic representations of chemical reactions.

The reactants are written on the left side whereas the products are written on the right side of an arrow.

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

The symbols for elements are written on the basis of their atomicity.

Atomicity is defined as the number of atoms in a molecule of an element.

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

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

A balanced chemical equation shows the same number of atoms of each element on both sides of the arrow.

Chemical equations are balanced on the basis of the Law of Conservation of Mass.

Answer to Problem 25QP

Solution:

a) ${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\to {\text{ 2N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ + O}}_{\text{2}}$

b) ${\text{2KNO}}_3\to {\text{ 2KNO}}_2{\text{ + O}}_{\text{2}}$

c) ${\text{NH}}_4{\text{NO}}_{\text{3}}\to {\text{ N}}_{\text{2}}{\text{O + 2H}}_{\text{2}}\text{O}$

d) ${\text{NH}}_4{\text{NO}}_2\to {\text{ N}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

e) ${\text{2NaHCO}}_{\text{3}}\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

f) ${\text{P}}_{\text{4}}{\text{O}}_{10}{\text{ + 6H}}_{\text{2}}\text{O }\to {\text{ 4H}}_{\text{3}}{\text{PO}}_{\text{4}}$

g) ${\text{2HCl + CaCO}}_{\text{3}}\to {\text{ CaCl}}_{\text{2}}{\text{ + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

h) ${\text{2Al + H}}_{\text{2}}{\text{SO}}_4\to {\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{ + 3H}}_{\text{2}}$

i) ${\text{CO}}_{\text{2}}\text{ + 2KOH }\to {\text{ K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}$

j) ${\text{CH}}_4{\text{ + 2O}}_{\text{2}}\to {\text{ CO}}_2{\text{ + 2H}}_{\text{2}}\text{O}$

k) ${\text{Be}}_{\text{2}}{\text{C + 4H}}_{\text{2}}\text{O }\to {\text{ 2Be(OH)}}_2{\text{ + CH}}_{\text{4}}$

l) ${\text{3Cu + 8HNO}}_3\to {\text{ 3Cu(NO}}_{\text{3}}{\text{)}}_2{\text{ + 2NO + 4H}}_2\text{O}$

m) ${\text{S + 6HNO}}_3\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 6NO}}_{\text{2}}{\text{ + 2H}}_2\text{O}$

n) ${\text{2NH}}_{\text{3}}+\text{3CuO}\to \text{3Cu }+{\text{N}}_{\text{2}}+{\text{3H}}_2\text{O}$

Explanation of Solution

a) ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\to {\text{ N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ + O}}_{\text{2}}$

There are two atoms of nitrogen on either side of the equation. However, there are five oxygen atoms on the left side and six oxygen atoms on the right side of the equation. The entire equation is balanced by adding the coefficient $\text{2}$ in front of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$. The balanced equation is written as follows:

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\to {\text{ 2N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ + O}}_{\text{2}}$

b) ${\text{KNO}}_3\to {\text{ KNO}}_2{\text{ + O}}_{\text{2}}$

There is one atom of potassium on either side of the equation. Also, there is one atom of nitrogen on both the sides. However, on the left there are three oxygen atoms while on the right, there are four oxygen atoms. The entire equation is balanced by adding the coefficient $\text{2}$ in front of ${\text{KNO}}_{\text{3}}$ and ${\text{KNO}}_{\text{2}}$. The balanced equation is written as follows:

${\text{2KNO}}_3\to {\text{ 2KNO}}_2{\text{ + O}}_{\text{2}}$

c) ${\text{NH}}_4{\text{NO}}_{\text{3}}\to {\text{ N}}_{\text{2}}{\text{O + H}}_{\text{2}}\text{O}$

There are two atoms of nitrogen on either side of the equation. However, there is an imbalance in the number of hydrogen and oxygen atoms. In case of hydrogen, there are four atoms on the left side and two on the right side. In case of oxygen, there are three atoms on the left side and two on the right side. The entire equation is balanced by adding the coefficient $\text{2}$ in front of ${\text{H}}_{\text{2}}\text{O}$. The balanced equation is written as follows:

${\text{NH}}_4{\text{NO}}_{\text{3}}\to {\text{ N}}_{\text{2}}{\text{O + 2H}}_{\text{2}}\text{O}$

d) ${\text{NH}}_4{\text{NO}}_2\to {\text{ N}}_{\text{2}}{\text{+ H}}_{\text{2}}\text{O}$

There are two atoms of nitrogen on either side of the equation. However, there are four atoms of hydrogen on the left side and two atoms on the right side. Also, there are two oxygen atoms on the left side and one on the right side of the equation. The entire equation is balanced by adding the coefficient $\text{2}$ in front of ${\text{H}}_{\text{2}}\text{O}$. The balanced equation is written as follows:

${\text{NH}}_4{\text{NO}}_2\to {\text{ N}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

e) ${\text{NaHCO}}_{\text{3}}\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

There is one atom of sodium on the left side while two atoms on the right side of the equation. For hydrogen, there is one atom on the left but two atoms on the right. For carbon also, there is one atom in the left but two atoms on the right. For oxygen, there are three atoms on the left but six atoms on the right of the equation. The entire equation is balanced by adding the coefficient $\text{2}$ in front of ${\text{NaHCO}}_{\text{3}}$. The balanced equation is written as follows:

${\text{2NaHCO}}_{\text{3}}\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

f) ${\text{P}}_{\text{4}}{\text{O}}_{10}{\text{ + H}}_{\text{2}}\text{O }\to {\text{ H}}_{\text{3}}{\text{PO}}_{\text{4}}$

There are four atoms of phosphorus on the left side but only one on the right side of the equation. In case of hydrogen, there are two atoms on the left side and three on the right. In case of oxygen, there are eleven atoms on the left but four on the right. The entire equation is balanced by adding the coefficient $6$ in front of ${\text{H}}_{\text{2}}\text{O}$ and $4$ in front of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$. The balanced equation is written as follows:

${\text{P}}_{\text{4}}{\text{O}}_{10}{\text{ + 6H}}_{\text{2}}\text{O }\to {\text{ 4H}}_{\text{3}}{\text{PO}}_{\text{4}}$

g) ${\text{HCl + CaCO}}_{\text{3}}\to {\text{ CaCl}}_{\text{2}}{\text{ + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

There is only one atom of calcium on either side of the equation. It is the same case with the elements carbon and oxygen, where it is one carbon atom and three oxygen atoms on the either side of the equation. However, there are discrepancies in the number of hydrogen and chlorine atoms. In case of chlorine, there is one on the left side and two on the right side. In case of hydrogen, there is one atom on the left side but two on the right side. The entire equation is balanced by adding the coefficient $\text{2}$ in front of $\text{HCl}$. The balanced equation is written as follows:

${\text{2HCl + CaCO}}_{\text{3}}\to {\text{ CaCl}}_{\text{2}}{\text{ + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

h) ${\text{Al + H}}_{\text{2}}{\text{SO}}_4\to {\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{ + H}}_{\text{2}}$

Apart from hydrogen, the number of all other atoms is different on the either side of the equation. In case of hydrogen, there are two atoms on both sides of the equation. On the other hand, for aluminium, there is one atom on the left and two atoms on the right. For sulphur, there is one atom on the left but three on the right. For oxygen, there are four atoms on the left but twelve on the right. The entire equation is balanced by adding the coefficient $2$ in front of $\text{Al}$, $3$ in front of ${\text{H}}_{\text{2}}{\text{SO}}_4$ and $3$ in front of ${\text{H}}_{\text{2}}$. The balanced equation is written as follows:

${\text{2Al + 3H}}_{\text{2}}{\text{SO}}_4\to {\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{ + 3H}}_{\text{2}}$

i) ${\text{CO}}_{\text{2}}\text{ + KOH }\to {\text{ K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}$

There is one atom of carbon on both sides of the equation. For potassium, there is one atom on the left but two on the right side of the equation. For hydrogen, there is one atom on the left side and two on the right side of the equation. For oxygen, there is one atom on the left side but four on the right side. The entire equation is balanced by adding the coefficient $2$ in front of $\text{KOH}$. The balanced equation is written as follows:

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

j) ${\text{CH}}_4{\text{ + 2O}}_{\text{2}}\to {\text{ CO}}_2{\text{ + H}}_{\text{2}}\text{O}$

There is one atom of carbon on both sides of the equation. For hydrogen, there are four atoms on the left side and two on the right side of the equation. For oxygen, there are two atoms on the left side but three on the right side. The entire equation is balanced by adding the coefficient $2$ in front of ${\text{O}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ each. The balanced equation is written as follows:

${\text{CH}}_4{\text{ + 2O}}_{\text{2}}\to {\text{ CO}}_2{\text{ + 2H}}_{\text{2}}\text{O}$

k) ${\text{Be}}_{\text{2}}{\text{C + H}}_{\text{2}}\text{O }\to {\text{ Be(OH)}}_2{\text{ + CH}}_{\text{4}}$

There is one atom of carbon on both sides of the equation. For hydrogen, there are two atoms on the left side but six on the right side of the equation. For oxygen, there is one atom on the left side but two on the right side. For beryllium, there is one atom on the left side but one on the right side. The entire equation is balanced by adding the coefficient $4$ in front of ${\text{H}}_{\text{2}}\text{O}$ and $2$ in front of ${\text{Be(OH)}}_2$. The balanced equation is written as follows:

${\text{Be}}_{\text{2}}{\text{C + 4H}}_{\text{2}}\text{O }\to {\text{ 2Be(OH)}}_2{\text{ + CH}}_{\text{4}}$

l) ${\text{Cu + HNO}}_3\to {\text{ Cu(NO}}_{\text{3}}{\text{)}}_2{\text{ + NO + H}}_2\text{O}$

There is one atom of copper on both sides of the equation. For oxygen, there are three atoms on the left side but eight on the right side of the equation. For hydrogen, there is one atom on the left side but two on the right side. For nitrogen, there is one atom on the left side but three on the right side. The entire equation is balanced by adding the coefficient $3$ in front of $\text{Cu}$, $8$ in front of ${\text{HNO}}_3$, $3$ in front of ${\text{Cu(NO}}_{\text{3}}{\text{)}}_2$, $2$ in front of $\text{NO}$, and $4$ in front of ${\text{H}}_{\text{2}}\text{O}$. The balanced equation is written as follows:

${\text{3Cu + 8HNO}}_3\to {\text{ 3Cu(NO}}_{\text{3}}{\text{)}}_2{\text{ + 2NO + 4H}}_2\text{O}$

m) ${\text{S + HNO}}_3\to {\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + NO}}_{\text{2}}{\text{ + H}}_2\text{O}$

There is one atom of sulphur on both sides of the equation. For oxygen, there are three atoms on the left side but seven on the right side of the equation. For hydrogen, there is one atom on the left side but four on the right side. For nitrogen, there is one atom on the left side and one on the right side. The entire equation is balanced by adding the coefficient $6$ in front of ${\text{HNO}}_{\text{3}}$, $6$ in front of ${\text{NO}}_2$ and $2$ in front of ${\text{H}}_{\text{2}}\text{O}$. The balanced equation is written as follows:

${\text{S + 6HNO}}_3\to {\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 6NO}}_{\text{2}}{\text{ + 2H}}_2\text{O}$

n) ${\text{NH}}_{\text{3}}\text{ + CuO }\to {\text{ Cu + N}}_{\text{2}}{\text{ + H}}_2\text{O}$

There is one atom of copper on both sides of the equation. It is the same for oxygen. For hydrogen, there are three atoms on the left side but two on the right side. For nitrogen, there is one atom on the left side but two on the right side. The entire equation is balanced by adding the coefficient $2$ in front of ${\text{NH}}_{\text{3}}$, $3$ in front of $\text{CuO}$, $3$ in front of $\text{Cu}$ and $3$ in front of ${\text{H}}_{\text{2}}\text{O}$. The balanced equation is written as follows:

${\text{2NH}}_{\text{3}}\text{ + 3CuO }\to {\text{ 3Cu + N}}_{\text{2}}{\text{ + 3H}}_2\text{O}$