# How do i balance this equation ?

Jul 20, 2017

Complete combustion of butane........? Well the products are ALWAYS $\text{carbon dioxide}$ and $\text{water}$.

#### Explanation:

And the normal rigmarole is to balance the carbons, as carbon dioxide, and the hydrogens as water, and then balance the oxygens......

${C}_{4} {H}_{10} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$ $\text{carbons balanced}$

${C}_{4} {H}_{10} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right)$ $\text{hydrogens balanced}$

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right)$ $\text{oxygens balanced}$

If you like you could double the entire equation to remove the half-integral coefficient in front of the dioxygen, i.e.

$2 {C}_{4} {H}_{10} \left(g\right) + 13 {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 10 {H}_{2} O \left(l\right)$

I have never seen the need to do so, in that I think the arithmetic is easier in the former case. And while I cannot have 1/2 a dioxygen molecule, I can certainly have half a mole of dioxygen molecules. It is up to you (and your prof) which you prefer. If we double the reaction, note that we must necessarily double the thermodynamic properties.

Note that this reaction follows the pattern of all stoichiometric equations. Garbage out equals garbage in, and MASS is conserved, and charge is conserved.

Can you do the corresponding combustion for $\text{pentane}$, ${C}_{5} {H}_{12}$? Try it out!