Chapter 3, Problem 21QP

Write an unbalanced equation to represent each of the following reactions: (a) potassium hydroxide and phosphoric acid react to form potassium phosphate and water; (b) zinc and silver chloride react to form zinc chloride and silver; (c) sodium hydrogen carbonate reacts to form sodium carbonate, water, and carbon dioxide; (d) ammonium nitrite reacts to form nitrogen and water; and (e) carbon dioxide and potassium hydroxide react to form potassium carbonate and water. (0 Balance equations (a)-(e).

Interpretation Introduction

Interpretation: An unbalanced equation that represents each of the given reactions and the balanced chemical equation for the same given reactions are to be written.

Concept introduction:

A chemical reaction uses a symbol to describe what occurs in a particular chemical reaction.

A chemical statement is represented in a chemical reaction.

Answer to Problem 21QP

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

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

c) ${\text{NaHCO}}_3\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

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

e) ${\text{CO}}_2+\text{KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$

f)

$\begin{array}{c}\text{3KOH}+{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{3H}}_{\text{2}}\text{O}\\ \text{Zn}+\text{2AgCl}\to {\text{ZnCl}}_{\text{2}}+\text{2Ag}\\ {\text{2NaHCO}}_3\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}\\ {\text{NH}}_4{\text{NO}}_{\text{2}}\to {\text{N}}_2+{\text{2H}}_{\text{2}}\text{O}\end{array}$

And

${\text{CO}}_2+\text{2KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

a)Potassium hydroxide and phosphoric acid react to form potassium phosphate and water.

Potassium hydroxide is represented as $\text{ KOH}$, phosphoric acid is represented as ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$, and potassium phosphate is represented as ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$. So, the unbalanced chemical equation for the reaction of potassium hydroxide and phosphoric acid to produce potassium phosphate and water is given as follows:

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

b) Zinc and silver chloride react to form zinc chloride and silver.

Zinc is represented by the symbol $\text{Zn,}$ silver chloride is represented as $\text{AgCl,}$ zinc chloride is represented as ${\text{ZnCl}}_{\text{2}}$, and silver is represented by the symbol $\text{Ag}$. The unbalanced chemical equation representing the reaction of zinc and silver chloride to produce zinc chloride and silver is given as follows:

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

c) Sodium hydrogen carbonate reacts to form sodium carbonate, water, and carbon dioxide.

Sodium hydrogen carbonate is represented as ${\text{NaHCO}}_3$, sodium carbonate is represented as ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$, water is represented as ${\text{H}}_{\text{2}}\text{O}$, and carbon dioxide is represented as ${\text{CO}}_{\text{2}}$. The unbalanced chemical equation representing the reaction of sodium hydrogen carbonate to produce sodium carbonate, water, and carbon dioxide is given as follows:

${\text{NaHCO}}_3\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

d) Ammonium nitrate reacts to form nitrogen and water.

Ammonium nitrate is represented as ${\text{NH}}_4{\text{NO}}_{\text{2}}$, nitrogen is represented by ${\text{N}}_2$, and water is represented as ${\text{H}}_{\text{2}}\text{O}$. The unbalanced chemical equation representing the reaction of ammonium nitrate to produce nitrogen and water is given as follows:

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

e) Carbon dioxide and potassium hydroxide react to form potassium carbonate and water.

Carbon dioxide is represented as ${\text{CO}}_2$, potassium hydroxide is represented as $\text{KOH}$, and potassium carbonate is represented as ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$. Water is represented as ${\text{H}}_{\text{2}}\text{O}$. The unbalanced chemical equation representing the reaction of magnesium and oxygen to produce magnesium iodide is given as follows:

${\text{CO}}_2\text{ + KOH}\to {\text{ K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}$

f) Balance all above equation

Balancing of chemical reactions is done on the basis of “The Law of Conservation of Mass.” The balanced equation representing the given reactions are as follows:

The equation of part (a) is as follows:

$\begin{array}{l}\text{KOH}+{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ \text{ 1}-\text{K}-3\\ \text{ 4}-\text{H}-2\\ \text{ 5}-\text{O}-5\end{array}$

As a first step, add a coefficient of 3 in front of potassium on the reactant side.

$\begin{array}{l}\text{3KOH}+{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ \text{ 3}-\text{K}-3\\ \text{ 4}-\text{H}-2\\ \text{ 5}-\text{O}-5\end{array}$

Now, add acoefficient of 3 in front of hydrogen on the product side.

$\begin{array}{l}\text{3KOH}+{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ \text{ 3}-\text{K}-3\\ \text{ 6}-\text{H}-6\\ \text{ 5}-\text{O}-5\end{array}$

Thus,

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

Hence, the first equation is balanced.

Now, the equation of part (b) is as follows:

$\begin{array}{l}\text{Zn}+\text{AgCl}\to {\text{ ZnCl}}_{\text{2}}+\text{Ag}\\ \text{ 1}-\text{Zn}-1\\ \text{ 1}-\text{Ag}-1\\ \text{ 1}-\text{Cl}-2\end{array}$

As a first step, add a coefficient of 2 in front of $\text{AgCl}$ on the reactant side.

$\begin{array}{l}\text{Zn}+\text{2AgCl}\to {\text{ZnCl}}_{\text{2}}+\text{Ag}\\ \text{ 1}-\text{Zn}-1\\ \text{ 2}-\text{Ag}-1\\ \text{ 2}-\text{Cl}-2\end{array}$

Now, add acoefficient of 2 in front of $\text{Ag}$ on the product side.

$\begin{array}{l}\text{Zn}+\text{2AgCl}\to {\text{ZnCl}}_{\text{2}}+\text{2Ag}\\ \text{ 1}-\text{Zn}-1\\ \text{ 2}-\text{Ag}-2\\ \text{ 2}-\text{Cl}-2\end{array}$

Thus,

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

Hence, the second equation is balanced.

Now, the equation of part (c) is as follows:

$\begin{array}{l}{\text{NaHCO}}_3\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}\\ \text{ 1}-\text{Na}-2\\ \text{ 1}-\text{H}-2\\ \text{ 1}-\text{C}-2\\ \text{ 3}-\text{O}-6\end{array}$

As a first step, add a coefficient of 2 in front of ${\text{NaHCO}}_3$ on the reactant side.

$\begin{array}{l}{\text{2NaHCO}}_3\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}\\ \text{ 2}-\text{Na}-2\\ \text{ 2}-\text{H}-2\\ \text{ 2}-\text{C}-2\\ \text{ 6}-\text{O}-6\end{array}$

Thus,

${\text{2NaHCO}}_3\to {\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

Hence, the third equation is balanced.

Now, the equation of part (d) is as follows:

$\begin{array}{l}{\text{NH}}_4{\text{NO}}_{\text{2}}\to {\text{ N}}_2+{\text{H}}_{\text{2}}\text{O}\\ \text{ 2}-\text{N}-2\\ \text{ 4}-\text{H}-2\\ \text{ 2}-\text{O}-1\end{array}$

As a first step, add a coefficient of 2 in front of ${\text{H}}_{\text{2}}\text{O}$ on the product side.

$\begin{array}{l}{\text{NH}}_4{\text{NO}}_{\text{2}}\to {\text{ N}}_2+2{\text{H}}_{\text{2}}\text{O}\\ \text{ 2}-\text{N}-2\\ \text{ 4}-\text{H}-4\\ \text{ 2}-\text{O}-2\end{array}$

Thus,

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

Hence, the fourth equation is balanced.

Now, the equation of part (e) is as follows:

$\begin{array}{l}{\text{CO}}_2\text{ + KOH}\to {\text{ K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}\\ \text{ 1}-\text{C}-1\\ \text{ 1}-\text{K}-2\\ \text{ 3}-\text{O}-4\\ \text{ 1}-\text{H}-2\end{array}$

As a first step, add a coefficient of 2 in front of $\text{KOH}$ on the reactant side.

$\begin{array}{l}{\text{CO}}_2\text{ + 2KOH}\to {\text{ K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}\\ \text{ 1}-\text{C}-1\\ \text{ 2}-\text{K}-2\\ \text{ 4}-\text{O}-4\\ \text{ 2}-\text{H}-2\end{array}$

Thus,

${\text{CO}}_2+\text{2KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$

Hence, the fifth equation is balanced.