Balancing Combustion Reactions PDF

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A guide and a few examples on how to Balance Combustion Reactions....

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Balancing Combustion Reactions

𝐶𝑥 𝐻𝑦 𝑂𝑧 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂 In a typical chemistry class, the combustion reaction that you will encounter usually has something to do with Hydrocarbon 𝐶𝑥 𝐻𝑦 and sometimes Oxygen might be present 𝐶𝑥 𝐻𝑦 𝑂𝑧 . The Hydrocarbon always reacts with Oxygen in a combustion reaction 𝑂2 .

𝐶𝑥 𝐻𝑦 𝑂𝑧 + 𝑂2 And the side products are always Carbon dioxide and Water, if it’s a complete combustion reaction.

𝐶𝑥 𝐻𝑦 𝑂𝑧 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂 But if you have a limited supply of Oxygen 𝑂2 , sometimes you might get other products such as Carbon Monoxide (CO) in addition with Carbon Dioxide.

𝐶𝑥 𝐻𝑦 𝑂𝑧 → 𝐶𝑂 + 𝐶𝑂2 But in a complete combustion, you will always have 𝐶𝑂2 𝑎𝑛𝑑 𝑤𝑎𝑡𝑒𝑟 𝐻2 𝑂. EXAMPLE 1. If we react Methane with excess oxygen. The products will be Carbon dioxide and water.

𝐶𝐻4 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂 STEPS: To balance it, start with Carbon first. Then Hydrogen and Oxygen.

𝐶𝐻4 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂 Carbon on reactant: 1

Carbon on product: 1

Then the Hydrogen. We have 4 on the left and 2 on the right. So put 2 at the beginning or the coefficient of 𝐻2 𝑂 to balance the Hydrogen atoms. (Note: You can only add to the coefficient but not to the subscript).

𝐶𝐻4 + 𝑂2 → 𝐶𝑂2 + 2𝐻2𝑂 Hydrogen on reactant: 4

Hydrogen on product: 4

So now we have 2 Oxygen atoms on the reactant and 4 Oxygen atoms on the product. Therefore, we need to put 2 o the 𝑂2 to balance it.

𝐶𝐻4 + 2𝑂2 → 𝐶𝑂2 + 2𝐻2 𝑂 Oxygen on reactant: 4

Oxygen on product: 4

EXAMPLE 2. Let’s react propane with Oxygen Gas. The product will also be Carbon dioxide and Water.

𝐶3𝐻8 + 𝑂2 → 𝐶𝑂2 + 𝐻2 𝑂 STEPS:

𝐶3𝐻8 + 𝑂2 → 3𝐶𝑂2 + 𝐻2𝑂 Carbon on reactant: 3

Carbon on product: 3

𝐶3𝐻8 + 𝑂2 → 3𝐶𝑂2 + 4𝐻2𝑂 Hydrogen on reactant: 8

Hydrogen on product: 8

Now we have 2 Oxygen atoms on the reactant and 10 Oxygen atoms on the product so:

𝐶3𝐻8 + 5𝑂2 → 3𝐶𝑂2 + 4𝐻2𝑂 Oxygen on reactant: 10

Oxygen on product: 10

EXAMPLE 3. React ethane with oxygen gas. The product will also be carbon dioxide and water.

𝐶2𝐻6 + 𝑂2 → 𝐶𝑂2 + 𝐻2 𝑂 STEPS:

𝐶2𝐻6 + 𝑂2 → 2𝐶𝑂2 + 𝐻2𝑂 Carbon on reactant: 2

Carbon on product: 2

𝐶2𝐻6 + 𝑂2 → 2𝐶𝑂2 + 3𝐻2 𝑂 Hydrogen on reactant: 6

Hydrogen on product: 6

So now we have 2 Oxygen atoms on the reactant and 7 Oxygen atoms on the product. To balance it:

7 𝐶2𝐻6 + 𝑂2 → 2𝐶𝑂2 + 3𝐻2 𝑂 2 But we can’t keep a fraction in the coefficient, so we need to multiply both sides (reactant and product) by the denominator which is 2, to eliminate the fraction. (Note: Only the coefficient will be multiplied and not the subscript) 7

2 (𝐶2𝐻6 + 𝑂2 → 2𝐶𝑂2 + 3𝐻2 𝑂) 2 2

2𝐶2𝐻6 + 7𝑂2 → 4𝐶𝑂2 + 6𝐻2 𝑂

EXAMPLE 3. React butane with oxygen gas. The product will also be carbon dioxide and water.

𝐶4𝐻10 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂 STEPS:

𝐶4𝐻10 + 𝑂2 → 4𝐶𝑂2 + 𝐻2 𝑂 Carbon on reactant: 4

Carbon on product: 4

𝐶4𝐻10 + 𝑂2 → 4𝐶𝑂2 + 5𝐻2 𝑂 Hydrogen on reactant: 10

Hydrogen on product: 10

The Oxyge atoms in the reactant is 2 and the Oxygen atoms in the product is 13. So:

𝐶4𝐻10 +

13 𝑂 → 4𝐶𝑂2 + 5𝐻2 𝑂 2 2

But we can’t keep a fraction in the coefficient, so we need to multiply both sides (reactant and product) by the denominator which is 2, to eliminate the fraction. (Note: Only the coefficient will be multiplied and not the subscript)

2 (𝐶4𝐻10 +

13 2

𝑂2 → 4𝐶𝑂2 + 5𝐻2 𝑂) 2

2𝐶4𝐻10 + 13𝑂2 → 8𝐶𝑂2 + 10𝐻2 𝑂...