Can you guys help me out on a couple of questions???

[zibb3r]

These are the questions:

Write balanced equations for the combustion of the following hydrocarbons and include states:

A) Ethane

B) Octane

C) Butane

Even if 1 person can just help with A and another with B that would be great!

LOVE YOU GUYS!!!!!!!!!!!!! :smitten: :smitten: :smitten:

[taiga]

A)
2C2H6 + 7O2 -> 4CO2 + 6H2O

B)
2C8H18 + 25O2 -> 16CO2 + 18H2O

C)
2C4H10 + 13O2 -> 8CO2 + 10H2O


I have to go out soon, but if no one has given a step by step process by the time i get back this arvo, I'll post it up That's if you don't figure it out yourself though

[zibb3r]

Thanks taiga!!!!! :smitten: :coolsmiley:

[pi]

So, combustion is the reaction with oxygen... I'm doing complete combustion for all these, but if it is incomplete, CO₂ will be replaced with either CO (carbon monoxide) or C (carbon).

A) 2C₂H₆ + 7O₂ --> 4CO₂ + 6H₂O

B) 2C₈H₁₈ + 25O₂ -> 16CO₂ + 18H₂O

C) 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

Required theory:
• For alkanes: C_nH_{2n + 2}
• In complete combustion for alkanes, the reactants are the alkane and oxygen gas (O₂), and the products are water (initially in a gaseous state - H₂)) and carbon dioxide gas (CO₂)
• In incomplete combustion, same reactants, but the CO₂ in the products may be replaced with either a C or a CO.
• Combustion (aka burning) is an exothermic reaction


EDIT: Beaten by taiga (but my answer looks better)

Moderator action: removed real name, sorry for the inconvenience

[vea]

You have to know that combustion of hydrocarbons involve oxygen in the reactants, heat as a catalyst and carbon dioxide and H2O as products. From there on you should be able to write out the unbalanced equation then balance it.

[zibb3r]

THANKS EVERYBODY!!!!  I actually understand it now!!!!!!!!!!!  YAY ME!!! :smitten:

[taiga]

So, combustion is the reaction with oxygen... I'm doing complete combustion for all these, but if it is incomplete, CO₂ will be replaced with either CO (carbon monoxide) or C (carbon).

A) 2C₂H₆ + 7O₂ --> 4CO₂ + 6H₂O

B) 2C₈H₁₈ + 25O₂ -> 16CO₂ + 18H₂O

C) 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

Required theory:
• For alkanes: C_nH_{2n + 2}
• In complete combustion for alkanes, the reactants are the alkane and oxygen gas (O₂), and the products are water (initially in a gaseous state - H₂)) and carbon dioxide gas (CO₂)
• In incomplete combustion, same reactants, but the CO₂ in the products may be replaced with either a C or a CO.
• Combustion (aka burning) is an exothermic reaction


EDIT: Beaten by taiga (but my answer looks better)

*post deleted*

But yes pi is right, definitely need to know about incomplete combustion and stuff. Though it is not huge on exams, schools may place it on SACs.

It takes some practice to understand the steps required to balance the equation, but once you get there it should be simple. Btw, remember to simplify the equation as you would in mathematics

Moderator action: removed real name, sorry for the inconvenience

[pi]

More tips for *cheating* the balancing of alkane complete combustion

1. For alkanes: C_nH_{2n + 2}
2. The alkane will always have a '2' as its 'coefficient'
3. The H₂O will always have '2n + 2' as its 'coefficient'
4. The CO₂ will always have '2n' as its coefficient
5. The O₂ will always have '(n) + (2n +2) - 1' = '3n + 1' as its coefficient

eg. Octane complete combustion:
n = 8 (rule 1)
Octane has 2 as coefficient (rule 2).
H₂O will have: 2n + 2 = 16 + 2 = 18 as coefficient (rule 3).
CO₂ will have: 2n = 16 as coefficient (rule 4).
O₂ will have: 3n + 1 = 24 + 1 = 25 as coefficient (rule 5)

Therefore:
2C8H18 + 25O2 -> 16CO2 + 18H2O


Just in case you prefer the mathsy way of doing it... (although I prefer knowing how to balance stuff out normally...)

Moderator action: removed real name, sorry for the inconvenience

[Greatness]

More tips for *cheating* the balancing of alkane complete combustion

1. For alkanes: C_nH_{2n + 2}
2. The alkane will always have a '2' as its 'coefficient'
3. The H₂O will always have '2n + 2' as its 'coefficient'
4. The CO₂ will always have '2n' as its coefficient
5. The O₂ will always have '(n) + (2n +2) - 1' = '3n + 1' as its coefficient

eg. Octane complete combustion:
n = 8 (rule 1)
Octane has 2 as coefficient (rule 2).
H₂O will have: 2n + 2 = 16 + 2 = 18 as coefficient (rule 3).
CO₂ will have: 2n = 16 as coefficient (rule 4).
O₂ will have: 3n + 1 = 24 + 1 = 25 as coefficient (rule 5)

Therefore:
2C8H18 + 25O2 -> 16CO2 + 18H2O


Just in case you prefer the mathsy way of doing it... (although I prefer knowing how to balance stuff out normally...)

I really like that tip/shortcut, good method for checking your answers.

Moderator action: removed real name, sorry for the inconvenience