Question

C8H18+O2=CO2+H20?

Answer 1

`2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O`

Explanation:

We begin by balancing the `C` and the `H` atoms first,

`C_8H_18 + O_2 -> 8CO_2 + 9H_2O`

Now, counting the number of `O` atoms on either side;
`"reactant side" -> 2`
`"product side "-> 25`

So, `C_8H_18 + 25/2O_2 -> 8CO_2 + 9H_2O`

`=>2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O`

Answer 2

Is there a question here? Of course you want to represent the BALANCED combustion equation.

`C_8H_18 +25/2O_2rarr 8CO_2 + 9H_2O`

Explanation:

The usual rigmarole in these combustion equations is to balance the carbons as carbon dioxide...then balance the hydrogens as water, and THEN balance the oxygens by adding in dioxygen.. And so ...

`C_8H_18 rarr 8CO_2 + H_2O` `"; carbons balanced"`

`C_8H_18 rarr 8CO_2 + 9H_2O` `"; hydrogens balanced"`

`C_8H_18 +25/2O_2rarr 8CO_2 + 9H_2O` `"; oxygen balanced"`

And thus finally, garbage out equals garbage in, and the equation is stoichiometrically balanced with respect to mass and charge. You can double the equation if you like in order to remove the half-integral oxygen coefficient. I have never seen the need to do so, especially as the arithmetic is easier in this form.

How many grams of dioxygen gas are needed for combustion of `100*g` of octane? How much carbon dioxide will be released; take that atmosphere! Of course these combustion equations provide rich fodder for examiners. They can ask for the VOLUMES of product and reactant given standard conditions.