Chapter 13, Problem 2E

Balance these chemical equations.

1. (a) ${\text{SO}}_2+{\text{ O}}_2\stackrel{}{\to }{\text{SO}}_3$
2. (b) ${\text{NH}}_3+{\text{O}}_2\stackrel{}{\to }{\text{N}}_2+{\text{H}}_2\text{O}$
3. (c) ${\text{C}}_4{\text{H}}_{10}+{\text{O}}_2\stackrel{}{\to }{\text{CO}}_2+{\text{ H}}_2\text{O}$
4. (d) ${\text{Pb(NO}}_3{\text{)}}_2\stackrel{}{\to }\text{PbO }+{\text{ NO}}_2+{\text{ O}}_2$
5. (e) $\text{Al}+{\text{Fe}}_3{\text{O}}_4\stackrel{}{\to }{\text{Al}}_2{\text{O}}_3+\text{Fe}$
6. (f) $\text{Sr}+{\text{H}}_2\text{O}\stackrel{}{\to }{\text{Sr(OH)}}_2+{\text{H}}_2$

To determine

Balance the given chemical equation: ${\text{SO}}_{\text{2}}+{\text{O}}_2\to {\text{SO}}_{\text{3}}$

Answer to Problem 2E

The balanced chemical equation is written as,

${\text{2SO}}_{\text{2}}+{\text{O}}_2\to 2{\text{SO}}_{\text{3}}$

Explanation of Solution

Given info: The chemical equation is given as,

${\text{SO}}_{\text{2}}+{\text{O}}_2\to {\text{SO}}_{\text{3}}$

Explanation:

In the balanced chemical equation, the number of same atom in the product and reactant side should be equal.

The given chemical equation is written as,

${\text{SO}}_{\text{2}}+{\text{O}}_2\to {\text{SO}}_{\text{3}}$

In the above reaction, there are four oxygen atoms on the reactant side and there are three oxygen atoms on the product side. If ${\text{O}}_{\text{2}}$ is multiplied by $\frac{1}{2}$ then the above equation will be balanced. But the coefficient in the balanced equation should be a whole number not a fraction.

Therefore, multiply ${\text{SO}}_2$ with $2$ and ${\text{SO}}_3$ with $2$. The balanced chemical equation is written as,

${\text{2SO}}_{\text{2}}+{\text{O}}_2\to 2{\text{SO}}_{\text{3}}$

Conclusion:

Therefore, the balanced chemical equation is written as,

${\text{2SO}}_{\text{2}}+{\text{O}}_2\to 2{\text{SO}}_{\text{3}}$

To determine

Balance the given chemical equation: ${\text{NH}}_3+{\text{O}}_2\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}$

Answer to Problem 2E

The balanced chemical equation is written as,

${\text{4NH}}_3+3{\text{O}}_2\to 2{\text{N}}_2+6{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given info: The chemical equation is given as,

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

Explanation:

In the balanced chemical equation, the number of same atom in the product and reactant side should be equal.

The given chemical equation is written as,

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

In the above reaction, there is one nitrogen atom on the reactant side and two nitrogen atoms on the product side. Therefore multiply ${\text{NH}}_{\text{3}}$ with $2$. The chemical equation is written as,

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

In the above reaction, there are six hydrogen atoms on the reactant side and two hydrogen atoms on the product side. Therefore multiply ${\text{H}}_2\text{O}$ with $3$. The chemical equation is written as,

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

In the above reaction, there are two oxygen atoms on the reactant side and three oxygen atoms on the product side. Therefore, multiply ${\text{O}}_{\text{2}}$ with $\frac{3}{2}$. The chemical equation is written as,

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

In the balanced equation, the coefficients should be a whole number not a fraction. Therefore, multiply the above equation with $2$. The balanced chemical equation is written as,

${\text{4NH}}_3+3{\text{O}}_2\to 2{\text{N}}_2+6{\text{H}}_{\text{2}}\text{O}$

Conclusion:

Therefore, the balanced chemical equation is written as,

${\text{4NH}}_3+3{\text{O}}_2\to 2{\text{N}}_2+6{\text{H}}_{\text{2}}\text{O}$

To determine

Balance the given chemical equation: ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

Answer to Problem 2E

The balanced chemical equation is written as,

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+13{\text{O}}_2\to 8{\text{CO}}_2+10{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given info: The chemical equation is given as,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

Explanation:

In the balanced chemical equation, the number of same atom in the product and reactant side should be equal.

The given chemical equation is written as,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are four carbon atoms on the reactant side and one carbon atom on the product side. Therefore multiply ${\text{CO}}_{\text{2}}$ with $4$. The chemical equation is written as,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+{\text{O}}_2\to 4{\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are ten hydrogen atoms on the reactant side and two hydrogen atoms on the product side. Therefore multiply ${\text{H}}_2\text{O}$ with $5$. The chemical equation is written as,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+{\text{O}}_2\to 4{\text{CO}}_2+5{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are two oxygen atoms on the reactant side and thirteen oxygen atoms on the product side. Therefore, multiply ${\text{O}}_{\text{2}}$ with $\frac{13}{2}$. The chemical equation is written as,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+\frac{13}{2}{\text{O}}_2\to 4{\text{CO}}_2+5{\text{H}}_{\text{2}}\text{O}$

In the balanced equation, the coefficients should be a whole number not a fraction. Therefore, multiply the above equation with $2$. The balanced chemical equation is written as,

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+13{\text{O}}_2\to 8{\text{CO}}_2+10{\text{H}}_{\text{2}}\text{O}$

Conclusion:

Therefore, the balanced chemical equation is written as,

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+13{\text{O}}_2\to 8{\text{CO}}_2+10{\text{H}}_{\text{2}}\text{O}$

To determine

Balance the given chemical equation: $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to \text{PbO}+{\text{NO}}_2+{\text{O}}_{\text{2}}$

Answer to Problem 2E

The balanced chemical equation is written as,

$\text{2Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to 2\text{PbO}+4{\text{NO}}_2+{\text{O}}_{\text{2}}$

Explanation of Solution

Given info: The chemical equation is given as,

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to \text{PbO}+{\text{NO}}_2+{\text{O}}_{\text{2}}$

Explanation:

In the balanced chemical equation, the number of same atom in the product and reactant side should be equal.

The given chemical equation is written as,

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to \text{PbO}+{\text{NO}}_2+{\text{O}}_{\text{2}}$

In the above equation there are two nitrogen atoms on the reactant side and one nitrogen atom on the product side. Therefore, multiply ${\text{NO}}_{\text{2}}$ with $2$. The chemical equation after balancing nitrogen atoms is written as,

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to \text{PbO}+2{\text{NO}}_2+{\text{O}}_{\text{2}}$

In the above reaction, there are six oxygen atoms on the reactant side and there are seven oxygen atoms on the product side. Therefore, multiply ${\text{O}}_{\text{2}}$ with $\frac{1}{2}$. The chemical equation is written as,

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to \text{PbO}+2{\text{NO}}_2+\frac{1}{2}{\text{O}}_{\text{2}}$

In the balanced equation, the coefficients should be a whole number not a fraction. Therefore, multiply the above equation with $2$. The balanced chemical equation is written as,

$\text{2Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to 2\text{PbO}+4{\text{NO}}_2+{\text{O}}_{\text{2}}$

Conclusion:

Therefore, the balanced chemical equation is written as,

$\text{2Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\to 2\text{PbO}+4{\text{NO}}_2+{\text{O}}_{\text{2}}$

To determine

Balance the given chemical equation: $\text{Al}+{\text{Fe}}_{\text{3}}{\text{O}}_4\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+\text{Fe}$

Answer to Problem 2E

The balanced chemical equation is written as,

$\text{8Al}+3{\text{Fe}}_{\text{3}}{\text{O}}_4\to 4{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+9\text{Fe}$

Explanation of Solution

Given info: The chemical equation is given as,

$\text{Al}+{\text{Fe}}_{\text{3}}{\text{O}}_4\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+\text{Fe}$

Explanation:

In the balanced chemical equation, the number of same atom in the product and reactant side should be equal.

The given chemical equation is written as,

$\text{Al}+{\text{Fe}}_{\text{3}}{\text{O}}_4\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+\text{Fe}$

In the above equation, there are four oxygen atoms on the reactant side and three oxygen atoms on product side. Therefore, multiply ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ with $3$ and ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ with $4$. The chemical equation after balancing the oxygen atoms on each side is written as,

$\text{Al}+3{\text{Fe}}_{\text{3}}{\text{O}}_4\to 4{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+\text{Fe}$

In the above equation, there is one aluminium atom on the reactant side and eight aluminium atoms on the product side. Therefore, multiply $\text{Al}$ with $8$. The chemical equation after balancing aluminium atom on each side is written as,

$\text{8Al}+3{\text{Fe}}_{\text{3}}{\text{O}}_4\to 4{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+\text{Fe}$

In the above equation, there are nine iron atoms on the reactant side and one iron atom on the product side. Therefore, multiply $\text{Fe}$ with $9$. The balanced chemical equation is written as,

$\text{8Al}+3{\text{Fe}}_{\text{3}}{\text{O}}_4\to 4{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+9\text{Fe}$

Conclusion:

Therefore, the balanced chemical equation is written as,

$\text{8Al}+3{\text{Fe}}_{\text{3}}{\text{O}}_4\to 4{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}+9\text{Fe}$

To determine

Balance the given chemical equation: $\text{Sr}+{\text{H}}_{\text{2}}\text{O}\to \text{Sr}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$

Answer to Problem 2E

The balanced chemical equation is written as,

$\text{Sr}+2{\text{H}}_{\text{2}}\text{O}\to \text{Sr}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$

Explanation of Solution

Given info: The chemical equation is given as,

$\text{Sr}+{\text{H}}_{\text{2}}\text{O}\to \text{Sr}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$

Explanation:

In the balanced chemical equation, the number of same atom in the product and reactant side should be equal.

The given chemical equation is written as,

$\text{Sr}+{\text{H}}_{\text{2}}\text{O}\to \text{Sr}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$

In the above reaction, there are two hydrogen atoms and one oxygen atom on the reactant side and there are four hydrogen atoms and two oxygen atoms on the product side. Therefore, multiply ${\text{H}}_2\text{O}$ with $2$ on the reactant side. The balanced chemical equation is written as,

$\text{Sr}+2{\text{H}}_{\text{2}}\text{O}\to \text{Sr}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$

Conclusion:

Therefore, the balanced chemical equation is written as,

$\text{Sr}+2{\text{H}}_{\text{2}}\text{O}\to \text{Sr}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$