Chapter 7, Problem 7.24QA

Interpretation Introduction

To write:

The complete and balanced chemical equation for the given incomplete reactions

Answer to Problem 7.24QA

Solution:

a. ${2C}_5{H}_{10 \left(l\right)}+ {15O}_{2 \left(g\right)} \to {10CO}_{2 \left(g\right)}+ 10H_2{O}_{\left(g\right)}$

b. ${2C}_6{H}_{14 \left(l\right)}+ {19O}_{2 \left(g\right)} \to 12{CO}_{2 \left(g\right)}+ {14H}_2{O}_{\left(g\right)}$

c. ${2C}_8{H}_{10 \left(l\right)}+ {21O}_{2 \left(g\right)} \to {16CO}_{2 \left(g\right)}+ 10H_2{O}_{\left(g\right)}$

d. $C_9{H}_{12 \left(g\right)}+ 12O_{2 \left(g\right)} \to {9CO}_{2 \left(g\right)}+ {6H}_2{O}_{\left(g\right)}$

Explanation of Solution

First, we write the complete and balanced chemical equation describing complete combustion. Combustion involves rapid reaction with $O_2$. The products of complete combustion of hydrocarbons are ${CO}_2$ and $H_{2}O$.The first step involves writing a reaction expression with single molecules or formula unit of the reactants and products. Then, we take the inventories of the number of atoms of each of the elements in the reaction mixture and we balance them starting with those that appear in only one reactant and product.

a. $C_5{H}_{10 \left(l\right)}+ O_{2 \left(g\right)} \to {CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $5C+10H+2O \to 1C+3O+ 2H$

There is only one $C$ atom on the right and five on left hand side of reaction arrow. To balance $C$, we place a coefficient of $5$ in front of ${CO}_2$ and recalculate distribution of atoms on both sides.

$C_5{H}_{10 \left(l\right)}+ O_{2 \left(g\right)} \to {5CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $5C+10H+2O \to 5C+11O+ 2H$

There are two $H$ atoms on the right and ten on left hand side of reaction arrow. To balance $H$, we place a coefficient of $5$ in front of $H_{2}O$ and recalculate distribution of atoms on both sides.

$C_5{H}_{10 \left(l\right)}+ O_{2 \left(g\right)} \to {5CO}_{2 \left(g\right)}+ 5H_2{O}_{\left(g\right)}$

Atoms: $5C+10H+2O \to 5C+15O+ 10H$

This balances the number of $C$ and $H$ atoms but leaves an odd number of $O$ atoms on the product side. The only source of $O$ atoms is $O_2$, so we need an even number of $O$ atoms on the right. To balance $O$ atoms, multiply all the coefficients in the expression by $2$.

${2C}_5{H}_{10 \left(l\right)}+ {2O}_{2 \left(g\right)} \to {10CO}_{2 \left(g\right)}+ 10H_2{O}_{\left(g\right)}$

Atoms: $10C+20H+4O \to 10C+30O+ 20H$

We can now balance the number of $O$ atoms by replacing the coefficient $2$ in front of $O_2$ with $15$, which gives us $30 O$ atoms and a balanced equation.

${2C}_5{H}_{10 \left(l\right)}+ {15O}_{2 \left(g\right)} \to {10CO}_{2 \left(g\right)}+ 10H_2{O}_{\left(g\right)}$

Atoms: $10C+20H+30O \to 10C+30O+ 20H$

b. $C_6{H}_{14 \left(l\right)}+ O_{2 \left(g\right)} \to {CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $6C+14H+2O \to 1C+3O+ 2H$

There is only one $C$ atom on the right and six on left hand side of reaction arrow. To balance $C$, we place a coefficient of $6$ in front of ${CO}_2$ and recalculate distribution of atoms on both sides.

$C_6{H}_{14 \left(l\right)}+ O_{2 \left(g\right)} \to 6{CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $6C+14H+2O \to 6C+13O+ 2H$

There are two $H$ atoms on the right and $14$ on left hand side of reaction arrow. To balance $H$, we place a coefficient of $7$ in front of $H_{2}O$ and recalculate distribution of atoms on both sides.

$C_6{H}_{14 \left(l\right)}+ O_{2 \left(g\right)} \to 6{CO}_{2 \left(g\right)}+ {7H}_2{O}_{\left(g\right)}$

Atoms: $6C+14H+2O \to 6C+19O+ 14H$

This balances the number of $C$ and $H$ atoms but leaves an odd number of $O$ atoms on the product side. The only source of $O$ atoms is $O_2$, so we need an even number of $O$ atoms on the right. To balance $O$ atoms, multiply all the coefficients in the expression by $2$.

${2C}_6{H}_{14 \left(l\right)}+ {2O}_{2 \left(g\right)} \to 12{CO}_{2 \left(g\right)}+ {14H}_2{O}_{\left(g\right)}$

Atoms: $12C+28H+4O \to 12C+38O+ 28H$

We can now balance the number of $O$ atoms by replacing the coefficient $2$ in front of $O_2$ with $19$, which gives us $30 O$ atoms and a balanced equation.

${2C}_6{H}_{14 \left(l\right)}+ {19O}_{2 \left(g\right)} \to 12{CO}_{2 \left(g\right)}+ {14H}_2{O}_{\left(g\right)}$

Atoms: $12C+28H+38O \to 12C+38O+ 28H$

c. $C_8{H}_{10 \left(l\right)}+ O_{2 \left(g\right)} \to {CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $8C+10H+2O \to 1C+3O+ 2H$

There is only one $C$ atom on the right and eight on left hand side of reaction arrow. To balance $C$, we place a coefficient of $8$ in front of ${CO}_2$ and recalculate distribution of atoms on both sides.

$C_8{H}_{10 \left(l\right)}+ O_{2 \left(g\right)} \to {8CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $8C+10H+2O \to 8C+17O+ 2H$

There are two $H$ atoms on the right and ten on left hand side of reaction arrow. To balance $H$, we place a coefficient of $5$ in front of $H_{2}O$ and recalculate distribution of atoms on both sides.

$C_8{H}_{10 \left(l\right)}+ O_{2 \left(g\right)} \to {8CO}_{2 \left(g\right)}+ 5H_2{O}_{\left(g\right)}$

Atoms: $8C+10H+2O \to 8C+21O+ 10H$

This balances the number of $C$ and $H$ atoms but leaves an odd number of $O$ atoms on the product side. The only source of $O$ atoms is $O_2$, so we need an even number of $O$ atoms on the right. To balance $O$ atoms, multiply all the coefficients in the expression by $2$.

${2C}_8{H}_{10 \left(l\right)}+ {2O}_{2 \left(g\right)} \to {16CO}_{2 \left(g\right)}+ 10H_2{O}_{\left(g\right)}$

Atoms: $16C+20H+4O \to 16C+42O+ 20H$

We can now balance the number of $O$ atoms by replacing the coefficient $2$ in front of $O_2$ with $21$, which gives us $40 O$ atoms and a balanced equation.

${2C}_8{H}_{10 \left(l\right)}+ {21O}_{2 \left(g\right)} \to {16CO}_{2 \left(g\right)}+ 10H_2{O}_{\left(g\right)}$

Atoms: $16C+20H+42O \to 16C+42O+ 20H$

d. $C_9{H}_{12 \left(g\right)}+ O_{2 \left(g\right)} \to {CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $9C+12H+2O \to 1C+3O+ 2H$

There is only one $C$ atom on the right and three on left hand side of reaction arrow. To balance $C$, we place a coefficient of 9 in front of ${CO}_2$ and recalculate distribution of atoms on both sides.

$C_9{H}_{12 \left(g\right)}+ O_{2 \left(g\right)} \to {9CO}_{2 \left(g\right)}+ H_2{O}_{\left(g\right)}$

Atoms: $9C+12H+2O \to 9C+19O+ 2H$

There are two $H$ atoms on the right and eight on left hand side of reaction arrow. To balance $H$

 , we place a coefficient of $6$ in front of $H_{2}O$ and recalculate distribution of atoms on both sides.

$C_9{H}_{12 \left(g\right)}+ O_{2 \left(g\right)} \to {9CO}_{2 \left(g\right)}+ {6H}_2{O}_{\left(g\right)}$

Atoms: $9C+12H+2O \to 9C+24O+ 12H$

There are $24O$ atoms on the right and two on left hand side of reaction arrow. To balance $O$, we place a coefficient of $12$ in front of $O_2$ and recalculate distribution of atoms on both sides.

$C_9{H}_{12 \left(g\right)}+ 12O_{2 \left(g\right)} \to {9CO}_{2 \left(g\right)}+ {6H}_2{O}_{\left(g\right)}$

Atoms: $9C+12H+24O \to 9C+24O+ 12H$

Both sides of the reaction have the same number of each element. Therefore, this reaction is balanced.

Conclusion:

Balancing each atom on both sides of reaction arrow balances the entire reaction.