What is the balanced equation for the complete combustion of heptane C7H16

Balanced Equation for the Complete Combustion of Heptane

During the complete combustion of heptane $\text{C}_7\text{H}_{16}$, heptane reacts with oxygen ($\text{O}_2$) to form carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$).

The balanced chemical equation for this reaction can be written as:

Where:

• $\text{C}_7\text{H}_{16}$ represents heptane
• $\text{O}_2$ represents oxygen gas
• $\text{CO}_2$ represents carbon dioxide
• $\text{H}_2\text{O}$ represents water

Detailed Explanation

1. Identify reactants and products:

   • Reactants: Heptane ($\text{C}_7\text{H}_{16}$) and oxygen ($\text{O}_2$)
   • Products: Carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$)

2. Balance carbon atoms:

   Heptane has 7 carbon atoms, hence:

   $$\text{C}_7\text{H}_{16} \rightarrow 7 \, \text{CO}_2$$

3. Balance hydrogen atoms:

   Heptane has 16 hydrogen atoms, hence:

   $$\text{C}_7\text{H}_{16} \rightarrow 8 \, \text{H}_2\text{O}$$

4. Balance oxygen atoms:

   • From 7 $\text{CO}_2$, we get $7 \times 2 = 14$ oxygen atoms.
   • From 8 $\text{H}_2\text{O}$, we get $8 \times 1 = 8$ oxygen atoms.

   In total, we need $14 + 8 = 22$ oxygen atoms.

   Since oxygen $\text{O}_2$ is diatomic, we divide by 2:

   $$\frac{22 \text{ O}}{2} = 11 \, \text{O}_2$$

The final balanced equation:

By following these steps, we ensure that the number of atoms of each element is conserved in the reaction.

Explanation of the 1. Law of Conservation of Mass

The law of conservation of mass is one of the fundamental principles in chemistry and physics. It states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.

In other words, the total mass of the reactants in a chemical reaction equals the total mass of the products. This principle was formulated by Antoine Lavoisier in 1789, and it laid the foundation for modern chemistry.

Mathematical Expression

The law of conservation of mass can be mathematically represented as:

Example in a Chemical Reaction

Consider a simple chemical reaction where hydrogen gas reacts with oxygen gas to form water:

• Reactants: 2 molecules of $H_2$ and 1 molecule of $O_2$
• Products: 2 molecules of $H_2O$

If we calculate the total mass on both sides of the reaction, we find they are equal, illustrating the conservation of mass.

Practical Implications

1. Balancing Chemical Equations: Ensuring mass is conserved helps chemists balance chemical equations correctly.
2. Measurement and Experimentation: The principle is crucial for precise measurements in lab settings to ensure that no mass appears or disappears unexpectedly.
3. Closed Systems: In real-world applications, maintaining a closed system is critical to observe the law strictly, as open systems might exchange mass with the environment.

Understanding the law of conservation of mass is essential not only in theoretical studies but also in practical applications in various scientific and engineering disciplines.

Explanation

Balancing chemical equations is a fundamental task in chemistry that ensures the law of conservation of mass is met. This law states that atoms are neither created nor destroyed in a chemical reaction. Therefore, the number of atoms of each element must be the same on both sides of the equation.

Steps for Balancing Chemical Equations

1. Write the unbalanced equation: Begin by writing the correct formulas for all reactants and products. For example:

   $$\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$$

2. Count the number of atoms of each element on both sides of the equation.

3. Use coefficients to balance the atoms: Adjust the coefficients (the numbers in front of the molecules) to get the same number of each type of atom on both sides. For instance:

   $$\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$$

   Notice that there are 2 hydrogen atoms on the left and 2 on the right, but there are 2 oxygen atoms on the left and only 1 on the right.

4. Adjust coefficients methodically: Start by balancing the atoms that appear in only one reactant and one product. Here we balance the oxygen atoms:

   $$\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

   Now we have 2 oxygen atoms on both sides, but 4 hydrogen atoms on the right.

5. Re-balance if necessary: In this case, we adjust the hydrogen atoms:

   $$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

Checking Your Work

To ensure the equation is balanced, recount the number of each type of atom:

• Reactants: $2 \text{H}_2$ = 4 H atoms, $1 \text{O}_2$ = 2 O atoms
• Products: $2 \text{H}_2\text{O}$ = 4 H atoms, 2 O atoms

Both sides have 4 hydrogen atoms and 2 oxygen atoms, confirming that the equation is balanced.

Practical Example

Consider a more complex equation:

Step-by-step balancing:

1. Balanced C atoms first:

   $$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}$$

2. Balanced H atoms next:

   $$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$

3. Finally, balance O atoms:

   $$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$

After these adjustments, the equation is balanced.

Balancing demands practice, but once mastered, it provides clarity and accuracy in conveying chemical reactions.