Chapter 13, Problem 1E

(a)

To determine

Balance the given chemical equation: ${\text{CuCl}}_{\text{2}}+{\text{H}}_{\text{2}}\to \text{Cu}+\text{HCl}$

Answer to Problem 1E

The balanced chemical equation is written as,

${\text{CuCl}}_{\text{2}}+{\text{H}}_{\text{2}}\to \text{Cu}+2\text{HCl}$

Explanation of Solution

Given info: The chemical equation is given as,

${\text{CuCl}}_{\text{2}}+{\text{H}}_{\text{2}}\to \text{Cu}+\text{HCl}$

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,

${\text{CuCl}}_{\text{2}}+{\text{H}}_{\text{2}}\to \text{Cu}+\text{HCl}$

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

${\text{CuCl}}_{\text{2}}+{\text{H}}_{\text{2}}\to \text{Cu}+2\text{HCl}$

Conclusion:

Therefore, the balanced chemical equation is written as,

${\text{CuCl}}_{\text{2}}+{\text{H}}_{\text{2}}\to \text{Cu}+2\text{HCl}$

(b)

To determine

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

Answer to Problem 1E

The balanced chemical equation is written as,

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

Explanation of Solution

Given info: The chemical equation is given as,

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

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,

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

In the above reaction, there is one iron atom and two oxygen atoms in the reactant side and there are two iron atoms and three oxygen atoms on the product side. If iron on the reactant side is multiplied by $2$ then ${\text{O}}_{\text{2}}$ must be multiplied by $\frac{3}{2}$ to balance the chemical equation. But in the balanced equation coefficients should be a whole number not a fraction.

Therefore, multiply $\text{Fe}$ with $4$ and ${\text{O}}_{\text{2}}$ with $3$ on the reactant side. Multiply ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ with $2$. The balanced chemical equation is written as,

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

Conclusion:

Therefore, the balanced chemical equation is written as,

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

(c)

To determine

Balance the given chemical equation: $\text{Al}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}+{\text{H}}_{\text{2}}$

Answer to Problem 1E

The balanced chemical equation is written as,

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

Explanation of Solution

Given info: The chemical equation is given as,

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

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,

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

In the above reaction, there is one $\text{Al}$ atom on the reactant side and there are two $\text{Al}$ atoms on the product side. Therefore multiply $\text{Al}$ atom with $2$ on the reactant side. The chemical equation after balancing aluminium atom is written as,

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

In the above reaction, there is one ${\text{SO}}_{\text{4}}$ unit in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and there are three ${\text{SO}}_{\text{4}}$ units in ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$. Therefore, multiply ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ with $3$. The chemical equation after balancing ${\text{SO}}_{\text{4}}$ unit is written as,

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

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

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

Conclusion:

Therefore, the balanced chemical equation is written as,

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

(d)

To determine

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

Answer to Problem 1E

The balanced chemical equation is written as,

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

Explanation of Solution

Given info: The chemical equation is given as,

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

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,

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

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,

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

Conclusion:

Therefore, the balanced chemical equation is written as,

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

(e)

To determine

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

Answer to Problem 1E

The balanced chemical equation is written as,

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

Explanation of Solution

Given info: The chemical equation is given as,

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

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,

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

In the above reaction, there are three oxygen atoms on the reactant side and there are four 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 ${\text{KNO}}_{\text{3}}$ with $2$ and ${\text{KNO}}_{\text{2}}$ with $2$. The balanced chemical equation is written as,

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

Conclusion:

Therefore, the balanced chemical equation is written as,

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

(f)

To determine

Balance the given chemical equation: ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}+{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Answer to Problem 1E

The balanced chemical equation is written as,

${\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}+15{\text{O}}_{\text{2}}\to 12{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given info: The chemical equation is given as,

${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}+{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

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,

${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}+{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are six carbon atoms on the reactant side and one carbon atom on the product side. Therefore multiply ${\text{CO}}_{\text{2}}$ with $6$. The chemical equation is written as,

${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}+{\text{O}}_{\text{2}}\to 6{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

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,

${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}+{\text{O}}_{\text{2}}\to 6{\text{CO}}_{\text{2}}+3{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are two oxygen atoms on the reactant side and fifteen oxygen atoms on the product side. Therefore, multiply ${\text{O}}_{\text{2}}$ with $\frac{15}{2}$. The chemical equation is written as,

${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}+\frac{15}{2}{\text{O}}_{\text{2}}\to 6{\text{CO}}_{\text{2}}+3{\text{H}}_{\text{2}}\text{O}$

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,

${\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}+15{\text{O}}_{\text{2}}\to 12{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$

Conclusion:

Therefore, the balanced chemical equation is written as,

${\text{2C}}_{\text{6}}{\text{H}}_{\text{6}}+15{\text{O}}_{\text{2}}\to 12{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}$