How do you balance #CH_4 + O_2 -> CO_2 + H_2O#?

2 Answers
Mar 30, 2017

Answer:

#CH_4+2O_2->CO_2+2H_2O#

Explanation:

Look for imbalances. Both sides have one #C#. However, the left-hand side has four #H#'s while the right-hand side has only two. The left-hand side has two #O#'s while the right-hand side has three.

We try to add some coefficients before each molecule to balance the number of atoms. We add a #2# before #H_2O# in the right-hand side as this would balance the number of #H#'s. But the right-hand side has four #O#'s now. Luckily, if we add a #2# before #O_2# in the left-hand side, the #O#'s would balance out.

Mar 30, 2017

Answer:

Balance the #"carbons"#, then the #"hydrogens"#, and then LASTLY the #"oxygens......"#

Explanation:

And thus we follow the rigmarole for combustion of methane...

#underbrace(CH_4 + O_2 rarr CO_2 + H_2O)_"carbons are balanced"#; .

#underbrace(CH_4 + O_2 rarr CO_2 + 2H_2O)_"carbons and hydrogens are balanced"#;.

And then, #underbrace(CH_4 + 2O_2 rarr CO_2 + 2H_2O)_"carbons and hydrogens and oxygens are balanced"#. And this equation is then stoichiometrically balanced.

With alkanes with an ODD number of carbons, this approach gives integral coefficients. Try it out for propane and pentane. When you balance alkanes with EVEN numbers of carbons, this approach requires the use of half-integral oxygen coefficients. e.g. for ethane:

#H_3C-CH_3 + 7/2O_2 rarr 2CO_2 +3H_2O#

Alternatively you could double the entire equation to get rid of the half-integral coefficient:

#2H_3C-CH_3 + 7O_2 rarr 4CO_2 +6H_2O#

Try this out for butane and hexane.

Because it is easier arithmetically to address the former stoichiometry, I tend to favour the half-integral coefficient. Of course we cannot have half a molecule of dioxygen, but we can certainly have #16*g#, i.e. #"half a mole of dioxygen gas."#